Q.What amount of heat must be supplied to $2.0 \times 10^{-2}\ \text{kg}$ of nitrogen (at room temperature) to raise its temperature by $45\,^{\circ}\text{C}$ at constant pressure? (Molecular mass of $\text{N}_2 = 28$; $R = 8.3\ \text{J mol}^{-1}\ \text{K}^{-1}$.)

Showing the 12 most recent of 27 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Equal amounts of a diatomic ideal gas are contained in two separate cylinders, P and Q. Cylinder P has a movable piston, while cylinder Q has a fixed piston. Both gases are initially at 273 K, and the same amount of heat is supplied to each. If the gas in cylinder P shows a temperature rise of 20 K, then the increase in temperature of the gas in cylinder Q is (A) 38 K (B) 20 K (C) 42 K (D) 28 K

   ›Reveal solutionSolution

   Equal heat causes a larger temperature rise at constant volume than at constant pressure (since some heat at constant pressure goes into expansion work); for a diatomic gas the ratio is exactly $C_p/C_v = 7/5$, giving 28 K.

   Concept and Intuition

   At constant pressure, part of the supplied heat does work pushing the piston outward, so less heat is 'left over' to raise the gas's internal energy/temperature — the effective heat capacity is $C_p$ (larger). At constant volume, no work is done, so all the heat raises the internal energy — the effective heat capacity is $C_v$ (smaller). Since $C_v < C_p$, the same amount of heat produces a larger temperature rise at constant volume than at constant pressure.

   Step-by-Step Solution

   1. Cylinder P has a movable piston $\Rightarrow$ constant-pressure process: $Q = nC_p\Delta T_P$.
   2. Cylinder Q has a fixed piston $\Rightarrow$ constant-volume process: $Q = nC_v\Delta T_Q$.
   3. Both cylinders receive the same heat $Q$ and contain the same amount ($n$) of the same diatomic gas, so:

   $nC_p\Delta T_P = nC_v\Delta T_Q \Rightarrow \Delta T_Q = \frac{C_p}{C_v}\Delta T_P.$

   4. For a diatomic ideal gas, $C_v = \dfrac{5}{2}R$, $C_p = \dfrac{7}{2}R$, so $\dfrac{C_p}{C_v} = \dfrac{7}{5} = 1.4$.
   5. $\Delta T_Q = 1.4 \times 20\ \text{K} = 28\ \text{K}$.

   Common Mistakes

   • Inverting the ratio (using $C_v/C_p$ instead of $C_p/C_v$) — remember the constant-volume rise must be larger, since none of its heat is 'spent' on work.
   • Using the monatomic ratio ($5/3$) instead of the diatomic ratio ($7/5$) by forgetting the gas is diatomic.

   ✓Final answer

   The correct option is (D) — 28 K.

   ANSWER: D

2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.When some amount of heat energy is supplied to a monoatomic gas at constant pressure P, the volume of the gas is increased by 42%. If the same amount of heat is supplied to a rigid diatomic gas at constant pressure 2P, then the percentage increase in the volume of the diatomic gas is (A) 63% (B) 15% (C) 21% (D) 30%

   ›Reveal solutionSolution

   The key is to relate heat input to volume change via the molar specific heat at constant pressure and the ideal gas law. For the monatomic gas, a 42% volume increase at pressure P corresponds to a certain heat input; applying the same heat to a diatomic gas at pressure 2P yields a 15% volume increase, so the correct option is (B).

   Concept and Intuition

   When heat is supplied at constant pressure, the gas expands and does work. The amount of heat needed to raise the temperature by $\Delta T$ is $Q = n C_p \Delta T$, where $C_p$ is the molar specific heat at constant pressure. For an ideal gas, the volume change at constant pressure is directly proportional to the temperature change: $V \propto T$ (from $PV = nRT$). So the fractional volume increase $\frac{\Delta V}{V} = \frac{\Delta T}{T}$.

   We are given that for a monatomic gas at pressure $P$, the volume increases by 42% when a certain heat $Q$ is supplied. For a diatomic gas at pressure $2P$, the same $Q$ is supplied. We need the new percentage volume increase. The trick: different gases have different $C_p$ values, and the pressure difference affects the initial volume but not the fractional change (since $V \propto T$ at constant $P$). We’ll compare the temperature rises and then the volume changes.

   Step-by-step solution

   1. Recall molar specific heats

      For a monatomic ideal gas: $C_v = \frac{3}{2}R$, so $C_p = C_v + R = \frac{5}{2}R$.

      For a rigid diatomic ideal gas (no vibration): $C_v = \frac{5}{2}R$, so $C_p = \frac{7}{2}R$.

   2. Express the heat supplied for the monatomic case

      Let $n$ be the number of moles (same gas amount in both cases).

      Heat supplied: $Q = n C_{p,\text{mono}} \Delta T_1 = n \cdot \frac{5}{2}R \cdot \Delta T_1$.

      At constant pressure $P$, the volume change is $\frac{\Delta V_1}{V_1} = \frac{\Delta T_1}{T_1}$.

      Given $\frac{\Delta V_1}{V_1} = 42\% = 0.42$, so $\Delta T_1 = 0.42\, T_1$.

   3. Find $Q$ in terms of $T_1$

      Substitute $\Delta T_1$:

   $Q = n \cdot \frac{5}{2}R \cdot (0.42\, T_1) = n R T_1 \cdot \frac{5}{2} \cdot 0.42 = n R T_1 \cdot 1.05.$

   4. Apply the same heat to the diatomic gas at pressure $2P$ For the diatomic gas: $Q = n C_{p,\text{di}} \Delta T_2 = n \cdot \frac{7}{2}R \cdot \Delta T_2$. Set equal to the previous $Q$:

   $n \cdot \frac{7}{2}R \cdot \Delta T_2 = n R T_1 \cdot 1.05.$

   Cancel $nR$:

   $\frac{7}{2} \Delta T_2 = 1.05\, T_1 \quad \Rightarrow \quad \Delta T_2 = \frac{2}{7} \cdot 1.05\, T_1 = 0.3\, T_1.$

   5. Relate $\Delta T_2$ to the volume change

      For the diatomic gas at constant pressure $2P$, the initial state obeys $2P V_2 = n R T_2$.

      But we don’t know $T_2$ directly. However, the fractional volume increase at constant pressure is still $\frac{\Delta V_2}{V_2} = \frac{\Delta T_2}{T_2}$.

      We need $T_2$ in terms of $T_1$. The initial conditions: the monatomic gas had pressure $P$ and volume $V_1$ at temperature $T_1$. The diatomic gas has pressure $2P$ and some volume $V_2$ at temperature $T_2$. The problem doesn’t specify that the initial volumes or temperatures are the same, but we can assume the same number of moles $n$ and that the gases start at the same temperature? Actually, careful: The problem only says “when some amount of heat is supplied… at constant pressure P” and then “the same amount of heat is supplied to a rigid diatomic gas at constant pressure 2P”. It does not state that the initial temperatures are equal. However, the percentage increase in volume depends only on the ratio $\Delta T / T$ for the given initial state. We need to find $T_2$ relative to $T_1$ from the ideal gas law using the given pressures and the fact that the heat input is the same. But wait — the initial volumes are not given, so we must assume the initial temperatures are the same? Let’s check: If the gases are initially at the same temperature $T_0$, then for the monatomic gas: $P V_1 = n R T_0$. For the diatomic gas: $2P V_2 = n R T_0$, so $V_2 = V_1/2$. That is plausible. But does the problem imply that? Usually in such problems, the initial temperature is taken as the same unless stated otherwise. Let’s verify with the numbers: If $T_2 = T_1$, then $\Delta T_2 = 0.3 T_1$ gives $\frac{\Delta V_2}{V_2} = 0.3 = 30\%$. That would be option (D). But we got 0.3 from the equation, so is it 30%? Wait, we must check: The calculation gave $\Delta T_2 = 0.3 T_1$. If $T_2 = T_1$, then percentage increase = 30%. But the answer choices include 30% as (D). However, we need to be careful: The initial temperature of the diatomic gas might not be the same as that of the monatomic gas. Let’s re-examine.

      Actually, the problem does not specify initial temperatures. But we can deduce them from the fact that the same amount of heat is supplied. The heat $Q$ is fixed. For the monatomic gas, we expressed $Q$ in terms of $T_1$. For the diatomic gas, we have $Q = n \frac{7}{2}R \Delta T_2$. So $\Delta T_2 = \frac{2Q}{7nR}$. But we also have $Q = 1.05 n R T_1$, so $\Delta T_2 = \frac{2}{7} \cdot 1.05 T_1 = 0.3 T_1$. That is correct. Now, the fractional volume increase is $\Delta V_2 / V_2 = \Delta T_2 / T_2$. We need $T_2$. The initial state of the diatomic gas is at pressure $2P$ and some volume $V_2$. But we don’t know $V_2$ or $T_2$. However, we can relate $T_2$ to $T_1$ if we assume the gases are initially at the same temperature? That is a common hidden assumption in such problems. Let’s test: If $T_2 = T_1$, then answer is 30%. But let’s see if there’s another relation.

      Alternatively, perhaps the initial volumes are the same? No, that would give different pressures. The problem likely intends that the initial conditions (temperature and number of moles) are the same for both gases, only the pressure differs because the container is different. So we set $T_1 = T_2 = T$ (initial). Then:

   $\frac{\Delta V_2}{V_2} = \frac{0.3 T}{T} = 0.3 = 30\%.$

   That gives option (D). But wait, the answer choices include 15%, 21%, 30%, 63%. Let’s double-check the calculation: For monatomic, $C_p = 5R/2$, 42% volume increase means $\Delta T/T = 0.42$, so $Q = n (5R/2) (0.42 T) = nR T \cdot 1.05$. For diatomic, $C_p = 7R/2$, so $\Delta T = \frac{Q}{n (7R/2)} = \frac{1.05 n R T}{3.5 n R} = 0.3 T$. So if initial $T$ same, volume increase = 30%. That seems straightforward. But why is 15% an option? Perhaps I missed that the diatomic gas is rigid (no vibration) so $C_v = 5R/2$ is correct. Let’s check if the pressure difference affects the initial volume? The fractional change doesn’t depend on initial volume, only on temperature change. So 30% seems correct.

   However, let’s re-read the problem: “When some amount of heat energy is supplied to a monoatomic gas at constant pressure P, the volume of the gas is increased by 42%. If the same amount of heat is supplied to a rigid diatomic gas at constant pressure 2P, then the percentage increase in the volume of the diatomic gas is” — note that the pressures are different. Could it be that the initial volumes are the same? If the initial volumes are the same, then from $PV = nRT$, for monatomic: $P V = n R T_1$, for diatomic: $2P V = n R T_2$, so $T_2 = 2 T_1$. Then the fractional volume increase for diatomic would be $\Delta V_2 / V_2 = \Delta T_2 / T_2$. We have $\Delta T_2 = 0.3 T_1$ from the heat equation, so $\Delta V_2 / V_2 = (0.3 T_1) / (2 T_1) = 0.15 = 15\%$. That gives option (B). That makes sense: the initial temperature of the diatomic gas is higher because at the same volume, higher pressure means higher temperature. So the same temperature rise gives a smaller fractional increase. This is a classic pitfall: assuming initial temperatures are equal when pressures differ.

   Watch out

   A common mistake is to assume the initial temperatures are the same. But the problem states different constant pressures, and if the initial volumes are not specified, one might implicitly assume they are equal. However, the phrase “at constant pressure P” and “at constant pressure 2P” does not specify initial volumes. In such problems, it is typical to assume the same initial volume for both gases, because the heat is supplied to the gas in a container; the pressure is different because the container’s piston loading is different. So we should assume the same initial volume $V_0$. Let’s verify: If the initial volumes are the same, then the initial temperatures are different: $T_{\text{mono}} = \frac{P V_0}{nR}$, $T_{\text{di}} = \frac{2P V_0}{nR} = 2 T_{\text{mono}}$. That yields the 15% result.

   Let’s confirm with the numbers: For monatomic, $\Delta V_1 / V_0 = 0.42$, so $\Delta T_1 = 0.42 T_{\text{mono}}$. Then $Q = n (5R/2) (0.42 T_{\text{mono}}) = 1.05 n R T_{\text{mono}}$. For diatomic, $Q = n (7R/2) \Delta T_2$, so $\Delta T_2 = (2/7) * 1.05 T_{\text{mono}} = 0.3 T_{\text{mono}}$. Now, initial volume $V_0$ for diatomic, so $\Delta V_2 / V_0 = \Delta T_2 / T_{\text{di}} = (0.3 T_{\text{mono}}) / (2 T_{\text{mono}}) = 0.15 = 15\%$. So the answer is 15%.

   Tip

   When pressures differ, always check if initial volumes or temperatures are implicitly the same. Here, the natural assumption is same initial volume (the gas is in a container of fixed size before heating), leading to different initial temperatures.

   6. Final answer The percentage increase in volume for the diatomic gas is 15%.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.When an ideal diatomic gas is heated at constant pressure, the fraction of the heat utilised to increase the internal energy of the gas is (A) $\dfrac{2}{5}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   This tests the split of heat between internal energy and work at constant pressure — the fraction going into internal energy is always $C_v/C_p$. For a diatomic ideal gas this is 5/7.

   Concept and Intuition

   When a gas is heated at constant pressure, it must expand (since $P$ is fixed and $T$ rises, $V$ must rise by the ideal gas law). Some of the supplied heat raises the internal energy (temperature) of the gas, and the rest is spent doing work pushing back the surroundings as it expands. The internal energy rise for ANY process (not just constant pressure) is $\Delta U = nC_v\Delta T$, since $U$ depends only on $T$ for an ideal gas. But the heat SUPPLIED specifically at constant pressure is $Q = nC_p\Delta T$. So the fraction of heat that becomes internal energy is $C_v/C_p$, a fixed number for a given gas — for diatomic gases with 5 degrees of freedom (3 translational + 2 rotational), $C_v = \frac{5}{2}R$ and $C_p = C_v + R = \frac{7}{2}R$.

   Step-by-Step Solution

   1. At constant pressure, heat supplied: $Q = nC_p\Delta T$.
   2. Internal energy is a state function depending only on T: $\Delta U = nC_v\Delta T$ (true regardless of the path).
   3. Fraction of heat going to internal energy: $\dfrac{\Delta U}{Q} = \dfrac{C_v}{C_p}$.
   4. For a diatomic gas: $C_v = \dfrac{5}{2}R$, $C_p = \dfrac{7}{2}R$.
   5. Fraction $= \dfrac{5/2\,R}{7/2\,R} = \dfrac{5}{7}$.

   Common Mistakes

   • Confusing this with the fraction going to WORK, which would be $1 - 5/7 = 2/7$ (using $R/C_p$), not $2/5$.
   • Using the monatomic values ($C_v = 3/2R$) instead of diatomic ($C_v=5/2R$).

   ✓Final answer

   The correct option is (D) — $\dfrac{5}{7}$.

   ANSWER: D

4. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.At constant pressure, if the work done by a gas is 40% of the increase in the internal energy of the gas, then the specific heat capacity of the gas at constant volume is (Universal gas constant = 8.3 J mol$^{-1}$ K$^{-1}$) (A) 29.05 J mol$^{-1}$ K$^{-1}$ (B) 25.35 J mol$^{-1}$ K$^{-1}$ (C) 20.75 J mol$^{-1}$ K$^{-1}$ (D) 32.55 J mol$^{-1}$ K$^{-1}$

   ›Reveal solutionSolution

   Comparing the constant-pressure work ($nR\Delta T$) to the internal energy change ($nC_v\Delta T$) via the given 40% ratio pins down $C_v = 2.5R = 20.75\ \text{J mol}^{-1}\text{K}^{-1}$.

   Concept and Intuition

   At constant pressure, a gas expanding does work $W=P\Delta V=nR\Delta T$ (ideal gas law), while its internal energy changes by $\Delta U=nC_v\Delta T$ regardless of the process. The ratio $W/\Delta U = R/C_v$ is therefore a fixed property of the gas, independent of how much it's heated.

   Step-by-Step Solution

   1. $W = nR\Delta T$ (constant pressure); $\Delta U = nC_v\Delta T$.
   2. Given $W = 0.4\,\Delta U$: $nR\Delta T = 0.4\times nC_v\Delta T$.
   3. The $n\Delta T$ cancels: $R = 0.4\,C_v \Rightarrow C_v = \dfrac{R}{0.4} = 2.5R$.
   4. $C_v = 2.5\times8.3 = 20.75\ \text{J mol}^{-1}\text{K}^{-1}$.

   Common Mistakes

   • Confusing $C_v$ with $C_p$ in the ratio (the work here is compared to $\Delta U$, which uses $C_v$, not $C_p$).
   • Forgetting that $n$ and $\Delta T$ cancel out, and trying to solve for them separately using leftover given data.

   ✓Final answer

   The correct option is (C) — 20.75 J mol$^{-1}$ K$^{-1}$.

   ANSWER: C

5. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.When a polyatomic gas is heated at constant pressure, the percentage of heat given to the gas that is converted into external work is (Ratio of the specific heat capacities of the gas $= \dfrac{4}{3}$) (A) 30 (B) 25 (C) 20 (D) 45

   ›Reveal solutionSolution

   The fraction of heat converted to external work at constant pressure is the universal formula $(\gamma-1)/\gamma$; for a polyatomic gas with $\gamma = 4/3$ this comes out to exactly 25%.

   Concept and Intuition

   At constant pressure, the first law gives $Q = \Delta U + W$, where for an ideal gas $Q = nC_p\Delta T$, $\Delta U = nC_v\Delta T$, and the external work done by the expanding gas is $W = P\Delta V = nR\Delta T$ (ideal gas law at constant $P$). The fraction of the supplied heat that goes into external work is therefore

   $\frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p}.$

   Using $C_p - C_v = R$ and $\gamma = C_p/C_v$, one can show $C_p = \dfrac{\gamma R}{\gamma - 1}$, so $\dfrac{R}{C_p} = \dfrac{\gamma-1}{\gamma}$ — a clean, gas-independent formula in terms of $\gamma$ alone.

   Step-by-Step Solution

   1. Fraction of heat converted to work at constant pressure: $\dfrac{W}{Q} = \dfrac{R}{C_p} = \dfrac{\gamma-1}{\gamma}$.
   2. Substitute $\gamma = \dfrac43$: $\gamma - 1 = \dfrac13$.
   3. Fraction $= \dfrac{1/3}{4/3} = \dfrac13\times\dfrac34 = \dfrac14$.
   4. As a percentage: $\dfrac14\times100\% = 25\%$.

   Common Mistakes

   • Confusing this with the fraction of heat that becomes internal energy ($\Delta U/Q = 1/\gamma$, which would give 75% here, not what's asked).
   • Using $\gamma$ for a diatomic gas (7/5) instead of the given polyatomic value (4/3).

   ✓Final answer

   The correct option is (B) — 25.

   ANSWER: B

6. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.If the relation between the absolute temperature (T) and volume (V) of an ideal gas which expands adiabatically is $T \propto \dfrac{1}{\sqrt{V}}$, then the ratio of the specific heat capacities of the gas is (A) 1.3 (B) 1.5 (C) 1.4 (D) 2.0

   ›Reveal solutionSolution

   For an adiabatic process, matching the given $T$–$V$ power law to $TV^{\gamma-1}=$ const gives $\gamma=1.5$.

   Concept and Intuition

   For a reversible adiabatic process of an ideal gas, $PV^\gamma = \text{const}$. Using $PV=nRT$, this can be rewritten purely in terms of $T$ and $V$:

   $TV^{\gamma-1} = \text{const}$

   So whenever a problem gives you $T$ as some power of $V$ in an adiabatic process, you can read off $\gamma-1$ directly as (minus) that power.

   Step-by-Step Solution

   1. Given: $T \propto V^{-1/2}$, i.e. $T V^{1/2} = \text{const}$.
   2. Compare exponents with the adiabatic relation $T V^{\gamma - 1} = \text{const}$.
   3. Matching powers of $V$: $\gamma - 1 = \dfrac{1}{2}$.
   4. Hence $\gamma = 1 + \dfrac12 = 1.5$.

   Common Mistakes

   • Confusing the adiabatic relation between $T,V$ with the one between $P,V$ ($PV^\gamma=$ const) and mismatching the exponent sign.
   • Forgetting that $\gamma - 1$ is the magnitude of the power on $V$ in the $T$–$V$ form.

   ✓Final answer

   The correct option is (B) — 1.5.

   ANSWER: B

7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.If the differences between the specific heat capacities at constant pressure and constant volume of hydrogen and another gas are in the ratio 16 : 1, then the other gas is (A) nitrogen (B) oxygen (C) carbon (D) argon

   ›Reveal solutionSolution

   Since specific heat differences scale as $1/M$, a ratio of 16 between hydrogen and another gas means the other gas has 16 times hydrogen's molar mass — oxygen ($M=32$).

   Concept and Intuition

   Mayer's relation for one mole is $C_p - C_v = R$. But when working with specific heat capacities (per unit mass, i.e. $c_p = C_p/M$), dividing through by molar mass $M$ gives

   $c_p - c_v = \frac{R}{M}$

   So the difference between the specific heats is inversely proportional to the molar mass of the gas — a heavier gas has a smaller specific-heat difference.

   Step-by-Step Solution

   1. Write the specific heat difference for hydrogen and for the unknown gas:

   $(c_p-c_v)_{H_2} = \frac{R}{M_{H_2}}, \qquad (c_p-c_v)_{gas} = \frac{R}{M_{gas}}$

   2. Take the given ratio:

   $\frac{(c_p-c_v)_{H_2}}{(c_p-c_v)_{gas}} = \frac{M_{gas}}{M_{H_2}} = \frac{16}{1}$

   3. With $M_{H_2}=2\text{ g/mol}$:

   $M_{gas} = 16 \times 2 = 32\text{ g/mol}$

   4. A diatomic gas of molar mass 32 g/mol is oxygen, $O_2$.

   Common Mistakes

   • Inverting the ratio (forgetting that the specific-heat-difference ratio is inversely, not directly, proportional to the molar-mass ratio).
   • Using $M_{H_2}=1$ (atomic mass of H) instead of $2$ (molecular mass of $H_2$ gas).

   ✓Final answer

   The correct option is (B) — oxygen.

   ANSWER: B

8. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.The heat to be supplied to 20 g of oxygen to increase its temperature from 27 °C to 59 °C at constant pressure is (Universal gas constant = $8.3\ \mathrm{J\,mol^{-1}K^{-1}}$) (A) 1162 J (B) 423 J (C) 934 J (D) 581 J

   ›Reveal solutionSolution

   Oxygen is diatomic, so $C_p = \tfrac72R$; plugging in the given moles and temperature change gives $Q \approx 581$ J.

   Concept and Intuition

   At constant pressure, part of the heat supplied to a gas goes into internal energy (raising temperature) and part into work done during expansion — captured together by the molar heat capacity at constant pressure, $C_p$. For a diatomic ideal gas (like $O_2$), $C_p = \tfrac{7}{2}R$ (5 degrees of freedom: 3 translational + 2 rotational, plus the $R$ from the $pV=nRT$ work term).

   Step-by-Step Solution

   1. Moles of oxygen: $n = \dfrac{20\ \text{g}}{32\ \text{g/mol}} = 0.625$ mol.
   2. Temperature change: $\Delta T = 59 - 27 = 32$ K.
   3. Molar heat capacity at constant pressure (diatomic): $C_p = \dfrac72 R = 3.5 \times 8.3 = 29.05\ \mathrm{J\,mol^{-1}K^{-1}}$.
   4. Heat supplied: $Q = nC_p\Delta T = 0.625 \times 29.05 \times 32 \approx 581$ J.

   Common Mistakes

   • Using $C_v = \tfrac52R$ (constant volume) instead of $C_p = \tfrac72R$ since the process here is explicitly at constant pressure.
   • Using the wrong molar mass for oxygen (it's $O_2$, molar mass 32 g/mol, not atomic oxygen at 16 g/mol).

   ✓Final answer

   The correct option is (D) — 581 J.

   ANSWER: D

9. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.If four moles of hydrogen and two moles of helium form a gaseous mixture, then the molar specific heat capacity of the mixture at constant pressure is (A) $\dfrac{16R}{7}$ (B) $\dfrac{7R}{16}$ (C) $R$ (D) $\dfrac{19R}{6}$

   ›Reveal solutionSolution

   The molar $C_p$ of a gas mixture is the mole-weighted average of the individual $C_p$ values; combining diatomic $H_2$ ($7R/2$) and monatomic He ($5R/2$) in the given proportions gives $19R/6$.

   Concept and Intuition

   When two ideal gases are mixed (and don't react), the total internal energy and total enthalpy of the mixture are just the sums of the individual gases' contributions. Dividing by the total number of moles gives the mixture's molar heat capacities as mole-fraction-weighted averages of the pure-gas values.

   Step-by-Step Solution

   1. $H_2$ is diatomic: $C_{p,H_2} = \dfrac72R$. Moles $n_1 = 4$.
   2. He is monatomic: $C_{p,He} = \dfrac52R$. Moles $n_2 = 2$.
   3. Mixture's molar $C_p$:

   $C_{p,mix} = \frac{n_1C_{p,H_2} + n_2C_{p,He}}{n_1+n_2} = \frac{4\left(\frac72R\right) + 2\left(\frac52R\right)}{4+2} = \frac{14R+5R}{6} = \frac{19R}{6}$

   Common Mistakes

   • Simply averaging the two $C_p$ values without weighting by the number of moles of each gas.
   • Using $C_v$ values for one gas and $C_p$ for the other, or mixing up which gas is diatomic vs monatomic.

   ✓Final answer

   The correct option is (D) — $\dfrac{19R}{6}$.

   ANSWER: D

10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.When 80 J of heat is supplied to a gas at constant pressure, if the work done by the gas is 20 J, then the ratio of the specific heat capacities of the gas is (A) $\dfrac{4}{3}$ (B) $\dfrac{5}{3}$ (C) $\dfrac{7}{5}$ (D) $\dfrac{9}{7}$

    ›Reveal solutionSolution

    This tests extracting the ratio of specific heats $\gamma$ from the first law of thermodynamics at constant pressure; the answer is (A) $\dfrac{4}{3}$.

    Concept and Intuition

    At constant pressure, the heat supplied equals $nC_p\Delta T$, while the internal energy change (which depends only on temperature for an ideal gas) equals $nC_v\Delta T$ regardless of the process. The first law $Q = \Delta U + W$ then lets us find $\Delta U$ from the given $Q$ and $W$, and the ratio $Q/\Delta U$ directly gives $C_p/C_v = \gamma$.

    Step-by-Step Solution

    1. First law: $\Delta U = Q - W = 80 - 20 = 60$ J.
    2. $Q = nC_p\Delta T = 80$ J and $\Delta U = nC_v\Delta T = 60$ J.
    3. Ratio: $\gamma = \dfrac{C_p}{C_v} = \dfrac{nC_p\Delta T}{nC_v\Delta T} = \dfrac{Q}{\Delta U} = \dfrac{80}{60} = \dfrac{4}{3}$.

    Common Mistakes

    • Computing $Q/W$ instead of $Q/\Delta U$ for the ratio of specific heats.
    • Forgetting that $W$ here is the work done BY the gas, which subtracts from $Q$ to give $\Delta U$ (sign confusion).

    ✓Final answer

    The correct option is (A) — $\dfrac{4}{3}$.

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.The change in internal energy of given mass of a gas, when its volume changes from V to 3V at constant pressure P is ($\gamma$ - Ratio of the specific heat capacities of the gas) (A) $\dfrac{PV}{\gamma-1}$ (B) $\dfrac{2PV}{\gamma-1}$ (C) $\dfrac{3PV}{\gamma-1}$ (D) $\dfrac{PV}{2\gamma-1}$

    ›Reveal solutionSolution

    This tests the relation $\Delta U = \dfrac{P\Delta V}{\gamma-1}$ for an ideal gas undergoing an isobaric change; the answer is (B) $\dfrac{2PV}{\gamma-1}$.

    Concept and Intuition

    For an ideal gas, the change in internal energy depends only on the temperature change, $\Delta U = nC_v\Delta T$, regardless of the process. At constant pressure, the ideal gas law gives $P\Delta V = nR\Delta T$, which lets us trade $\Delta T$ for the more directly known quantity $\Delta V$. Combining these with $C_v = \dfrac{R}{\gamma - 1}$ gives a compact formula purely in terms of $P$ and $\Delta V$.

    Step-by-Step Solution

    1. Internal energy change: $\Delta U = nC_v\Delta T$.
    2. At constant pressure, $nR\Delta T = P\Delta V$, so $nC_v\Delta T = \dfrac{C_v}{R}(P\Delta V)$.
    3. Since $C_v = \dfrac{R}{\gamma-1}$, we get $\Delta U = \dfrac{P\Delta V}{\gamma - 1}$.
    4. Given $V \to 3V$ at constant $P$: $\Delta V = 3V - V = 2V$.
    5. Substituting: $\Delta U = \dfrac{P(2V)}{\gamma-1} = \dfrac{2PV}{\gamma-1}$.

    Common Mistakes

    • Forgetting that $\Delta U$ depends only on $\Delta T$ (state function), and trying to use $W = P\Delta V$ directly as $\Delta U$ (that's the work done, not the internal energy change).
    • Using $\Delta V = 3V$ instead of $2V$ (forgetting to subtract the initial volume).

    ✓Final answer

    The correct option is (B) — $\dfrac{2PV}{\gamma-1}$.

    ANSWER: B

12. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.Heat is supplied at constant pressure to a diatomic gas. The part of this heat, that was utilized to increase its internal energy is (A) $4/5$ (B) $5/7$ (C) $3/5$ (D) $5/6$

    ›Reveal solutionSolution

    This tests knowing the degrees-of-freedom-based specific heats of a diatomic gas ($C_v=\tfrac52R$, $C_p=\tfrac72R$) and using them to split constant-pressure heat between internal-energy increase and work done.

    Concept and Intuition

    At constant pressure, supplied heat $dQ$ splits into two parts: raising internal energy ($dU$) and doing work on the surroundings as the gas expands ($dW=PdV$), per the first law $dQ=dU+dW$. Since $dU=nC_vdT$ always (regardless of process) while $dQ=nC_pdT$ at constant pressure, the fraction of heat that goes into internal energy at constant pressure is simply the ratio $C_v/C_p$ — a fixed number depending only on the gas's degrees of freedom.

    Step-by-Step Solution

    1. For a diatomic ideal gas (5 degrees of freedom: 3 translational + 2 rotational), $C_v=\tfrac52R$ and $C_p=C_v+R=\tfrac72R$.
    2. At constant pressure, $dQ=nC_pdT$ and $dU=nC_vdT$, so the fraction going to internal energy is $\dfrac{dU}{dQ}=\dfrac{C_v}{C_p}$.
    3. Substituting: $\dfrac{C_v}{C_p}=\dfrac{5R/2}{7R/2}=\dfrac57$.

    Common Mistakes

    • Using the monoatomic values ($C_v=\tfrac32R$, giving $3/5$) instead of diatomic.
    • Confusing this ratio with $C_v/C_p=1/\gamma$ written the wrong way round.

    ✓Final answer

    The correct option is (B) — $5/7$.

    ANSWER: B