Q.Is it possible to increase the temperature of a gas without adding heat to it? Explain.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Two gases 'A' and 'B' are initially at same pressure, volume and temperature. If 'A' is compressed isothermally and 'B' adiabatically to half of the initial volume, then the final pressure of 'A' is (A) greater than the final pressure of 'B' (B) equal to the final pressure of 'B' (C) less than the final pressure of 'B' (D) twice the final pressure of 'B'

   ›Reveal solutionSolution

   This compares final pressures after equal-volume isothermal vs adiabatic compression, using $PV=\text{const}$ vs $PV^\gamma=\text{const}$. The answer is (C) less than the final pressure of B.

   Concept and Intuition

   An adiabatic curve is always steeper than an isothermal curve at any common point on a P-V diagram (since $\gamma>1$). This means for the same fractional decrease in volume, adiabatic compression raises the pressure by a larger factor than isothermal compression does. Physically, in adiabatic compression no heat escapes, so all the compression work raises the gas's internal energy (and hence temperature and pressure) further than in the isothermal case, where the gas can shed heat to keep temperature — and hence pressure rise — smaller.

   Step-by-Step Solution

   1. Both gases start at the same $P_0, V_0, T_0$.
   2. Gas A, isothermal to $V_0/2$: $P_0 V_0 = P_A (V_0/2) \Rightarrow P_A = 2P_0$.
   3. Gas B, adiabatic to $V_0/2$: $P_0 V_0^\gamma = P_B (V_0/2)^\gamma \Rightarrow P_B = P_0 \cdot 2^\gamma$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The pressure and density of a diatomic gas ($\gamma=\dfrac{7}{5}$) change adiabatically from (p, d) to $(p^1, d^1)$. If $\dfrac{d^1}{d}=32$, then $\dfrac{p^1}{p}$ should be (A) $\dfrac{1}{128}$ (B) 32 (C) 128 (D) $\dfrac{1}{32}$

   ›Reveal solutionSolution

   For an adiabatic change, pressure scales with density as $p\propto d^{\gamma}$; with $d'/d=32=2^5$ and $\gamma=7/5$, the exponent works out to a clean integer power, giving $p'/p=2^7=128$.

   Concept and Intuition

   The adiabatic condition $pV^{\gamma}=\text{constant}$ is usually written in terms of volume, but density $d=\dfrac{m}{V}$ is just $\propto 1/V$ for a fixed mass of gas. Substituting $V\propto 1/d$ turns the law into $p\,d^{-\gamma}=\text{constant}$, i.e. $p\propto d^{\gamma}$. This is a very handy restated form whenever a problem gives density ratios instead of volume ratios.

   Step-by-Step Solution

   1. Start from $pV^{\gamma} = p^1 V'^{\gamma}$.
   2. Replace $V\to \frac{1}{d}$ and $V'\to\frac{1}{d'}$: $p\,d^{-\gamma} = p^1 d'^{-\gamma}$.
   3. Rearrange: $\dfrac{p^1}{p} = \left(\dfrac{d^1}{d}\right)^{\gamma}$.
   4. Substitute $d^1/d=32$ and $\gamma=7/5$: $\dfrac{p^1}{p}=32^{7/5}$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The pressure $P_1$ and density $d_1$ of a diatomic gas change to $P_2$ and $d_2$ during an adiabatic operation. Find the value of $\dfrac{P_1}{P_2}$, if $\dfrac{d_2}{d_1} = 32$ (A) 128 (B) $\dfrac{1}{64}$ (C) 64 (D) $\dfrac{1}{128}$

   ›Reveal solutionSolution

   This tests relating pressure and density in an adiabatic process via $\gamma$ for a diatomic gas. Answer: $P_1/P_2 = 1/128$.

   Concept and Intuition

   In an adiabatic process, $PV^\gamma = \text{constant}$. Since density is inversely proportional to volume for a fixed mass of gas, this relation can be rewritten directly in terms of pressure and density: $P \propto \rho^\gamma$. For a diatomic gas, $\gamma = 7/5 = 1.4$.

   Step-by-Step Solution

   1. Adiabatic relation: $P_1V_1^\gamma = P_2V_2^\gamma \Rightarrow \dfrac{P_1}{P_2} = \left(\dfrac{V_2}{V_1}\right)^\gamma$.
   2. Since density $\rho = m/V$ (mass constant), $V \propto 1/\rho$, so $\dfrac{V_2}{V_1} = \dfrac{\rho_1}{\rho_2}$.
   3. Thus $\dfrac{P_1}{P_2} = \left(\dfrac{\rho_1}{\rho_2}\right)^\gamma = \left(\dfrac{d_1}{d_2}\right)^\gamma$.
   4. Given $\dfrac{d_2}{d_1} = 32$, so $\dfrac{d_1}{d_2} = \dfrac{1}{32}$. …

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4. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.During an adiabatic process, if the volume of 4 moles of a monoatomic gas initially at a temperature of $127^{\circ}\text{C}$ increases by 7 times, then the work done by the gas is ($R$ - universal gas constant) (A) 2400 R (B) 900 R (C) 1800 R (D) 1200 R

   ›Reveal solutionSolution

   Tests the adiabatic relation $TV^{\gamma-1}=\text{const}$ to find the temperature drop on expansion, then converts that into work done using $W=-\Delta U$ (since $Q=0$).

   Concept and Intuition

   In an adiabatic process no heat is exchanged, so any work the gas does on its surroundings comes entirely out of its internal energy — the gas must cool as it expands. Finding the final temperature from the adiabatic volume relation is the key step; after that, $W=nC_v(T_1-T_2)$ follows directly since $Q=0$.

   Step-by-Step Solution

   1. Initial temperature $T_1 = 127+273 = 400\,K$. The volume "increases by 7 times" means the increase itself equals $7V_1$, so the final volume is $V_2 = V_1+7V_1=8V_1$.
   2. Monoatomic gas: $\gamma = 5/3$, so $\gamma-1 = 2/3$.
   3. Adiabatic relation: $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\Rightarrow T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{2/3}=400\left(\dfrac18\right)^{2/3}$.
   4. $\left(\dfrac18\right)^{2/3}=\left(2^{-3}\right)^{2/3}=2^{-2}=\dfrac14$, so $T_2 = 400\times\dfrac14=100\,K$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If a gas of volume 400 cc at an initial pressure P is suddenly compressed to 100 cc, then its final pressure is (The ratio of the specific heat capacities of the gas at constant pressure and constant volume is 1.5) (A) $\dfrac{P}{32}$ (B) $8P$ (C) $32P$ (D) $16P$

   ›Reveal solutionSolution

   Sudden compression is adiabatic; applying $P_1V_1^\gamma = P_2V_2^\gamma$ with $\gamma=1.5$ and a volume ratio of 4 gives the final pressure $8P$.

   Concept and Intuition

   "Suddenly compressed" signals no time for heat exchange with the surroundings — this is an adiabatic process, governed by $PV^\gamma = \text{constant}$, not Boyle's law ($PV=\text{const}$, which is isothermal). The extra pressure rise beyond what Boyle's law would predict comes from the gas heating up as it's compressed (since no heat escapes, the compression work raises internal energy/temperature).

   Step-by-Step Solution

   1. Adiabatic relation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
   2. Given $V_1 = 400$ cc, $V_2=100$ cc, so $\dfrac{V_1}{V_2}=4$.
   3. $\gamma = C_p/C_v = 1.5$.
   4. $P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^{\gamma} = P\times 4^{1.5}$. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the given graph shows the logarithmic values of pressure (P) and volume (V) of an ideal gas, then the ratio of the specific heat capacities of the gas is [FIGURE] (a straight line graph of log P (Pa) on the y-axis vs log V (m^3) on the x-axis, sloping downward from left to right; the line passes through the points (1.2, 2.48) and (1.4, 2.20), each marked with dashed guide lines) (A) 1.5 (B) 1.2 (C) 1.4 (D) 1.3

   ›Reveal solutionSolution

   A straight line on a $\log P$ vs $\log V$ plot represents an adiabatic process, whose slope's magnitude directly gives $\gamma=C_P/C_V$. Answer: 1.4.

   Concept and Intuition

   For an adiabatic process of an ideal gas, $PV^\gamma=\text{constant}$. Taking logarithms: $\log P + \gamma\log V = \text{constant}$, i.e. $\log P = -\gamma \log V + \text{constant}$. This is a straight line with slope $-\gamma$ when plotted as $\log P$ (y-axis) versus $\log V$ (x-axis) — exactly the graph given, which decreases as $V$ increases (consistent with a negative slope).

   Step-by-Step Solution

   1. Slope of the line $= \dfrac{\Delta(\log P)}{\Delta(\log V)} = \dfrac{2.20-2.48}{1.4-1.2} = \dfrac{-0.28}{0.2} = -1.4$.
   2. Comparing to $\log P = -\gamma\log V + \text{const}$, the slope equals $-\gamma$.
   3. So $-\gamma = -1.4 \Rightarrow \gamma = 1.4$. …

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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If 5.6 litres of a monoatomic gas at STP is adiabatically compressed to 0.7 litres, then the work done on the gas is nearly (R - Universal gas constant) (A) 307R (B) 357R (C) 367R (D) 407R

   ›Reveal solutionSolution

   For an adiabatic compression, no heat is exchanged, so the work done on the gas equals its rise in internal energy; computing $T_2$ from $TV^{\gamma-1}=\text{const}$ and then $\Delta U$ gives about $307R$.

   Concept and Intuition

   In an adiabatic process, $Q=0$, so the first law of thermodynamics $\Delta U = Q - W_{\text{by gas}}$ reduces to $\Delta U = -W_{\text{by gas}} = W_{\text{on gas}}$. That is, all the work done on the gas during adiabatic compression goes directly into raising its internal energy — there is nowhere else for the energy to go since no heat can escape.

   For an adiabatic process, $TV^{\gamma-1} = \text{constant}$, which lets us find the final temperature purely from the volume ratio and $\gamma$ (here $\gamma=5/3$ for a monoatomic gas).

   Step-by-Step Solution

   1. Find moles at STP: $n = \dfrac{5.6\ \text{L}}{22.4\ \text{L/mol}} = 0.25$ mol; $T_1 = 273\ \mathrm K$.
   2. Volume ratio: $\dfrac{V_1}{V_2} = \dfrac{5.6}{0.7} = 8$.
   3. Adiabatic relation: $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = 273 \times 8^{2/3}$.
   4. $8^{2/3} = (2^3)^{2/3} = 2^2 = 4$, so $T_2 = 273\times4 = 1092\ \mathrm K$.
   5. $\Delta T = 1092 - 273 = 819\ \mathrm K$. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.When an ideal diatomic gas undergoes adiabatic expansion, if the increase in its volume is 0.5%, then the change in the pressure of the gas is (A) +0.5% (B) -0.5% (C) -0.7% (D) +0.7%

   ›Reveal solutionSolution

   Taking the logarithmic differential of the adiabatic law $PV^\gamma=\text{const}$ links fractional changes in pressure and volume through $\gamma$; for a diatomic gas ($\gamma=1.4$) a $+0.5\%$ volume change gives a $-0.7\%$ pressure change — option (C).

   Concept and Intuition

   In an adiabatic process, no heat is exchanged, and the gas obeys $PV^\gamma=\text{constant}$. For small (percentage-scale) changes, it's convenient to take the logarithmic derivative of this relation rather than solving it exactly — this converts the power-law relationship into a simple linear one between the fractional (percentage) changes in $P$ and $V$.

   Step-by-Step Solution

   1. Start from $PV^\gamma=\text{const}$.
   2. Take natural log: $\ln P+\gamma\ln V=\text{const}$.
   3. Differentiate: $\dfrac{dP}{P}+\gamma\dfrac{dV}{V}=0\ \Rightarrow\ \dfrac{dP}{P}=-\gamma\dfrac{dV}{V}$.
   4. For a diatomic ideal gas, $\gamma=\dfrac{7}{5}=1.4$. …

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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A monatomic gas at a pressure of 100 kPa expands adiabatically such that its final volume becomes 8 times its initial volume. If the work done during the process is 180 J, then the initial volume of the gas is (A) $1600\ \mathrm{cm^3}$ (B) $800\ \mathrm{cm^3}$ (C) $1200\ \mathrm{cm^3}$ (D) $2000\ \mathrm{cm^3}$

   ›Reveal solutionSolution

   This tests the adiabatic work formula $W=\dfrac{P_1V_1-P_2V_2}{\gamma-1}$ for a monatomic gas; the answer is (A) $1600\ \mathrm{cm^3}$.

   Concept and Intuition

   In an adiabatic process no heat is exchanged, so all the work done by the gas comes at the expense of its internal energy: $W = -\Delta U = \dfrac{P_1V_1-P_2V_2}{\gamma-1}$. For a monatomic gas $\gamma = 5/3$. Since we're given the volume ratio ($V_2=8V_1$), we can find $P_2$ in terms of $P_1$ using the adiabatic relation $PV^\gamma = \text{constant}$, and then everything reduces to one unknown, $V_1$.

   Step-by-Step Solution

   1. Given: $P_1 = 100\ \mathrm{kPa}$, $V_2 = 8V_1$, $W = 180\ \mathrm{J}$, monatomic so $\gamma = 5/3$.
   2. Adiabatic relation: $P_1V_1^\gamma = P_2V_2^\gamma \Rightarrow P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = P_1\left(\dfrac{1}{8}\right)^{5/3}$.
   3. Since $8 = 2^3$, $8^{5/3} = 2^5 = 32$, so $P_2 = \dfrac{P_1}{32}$.
   4. $P_2V_2 = \dfrac{P_1}{32}\times 8V_1 = \dfrac{P_1V_1}{4}$. …

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10. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.Two samples of same gas $\left(\gamma=\dfrac{3}{2}\right)$ have equal volume. If their volumes are doubled by adiabatic and isothermal processes respectively for sample 1 and 2. Their final pressures are now equal. The ratio of initial pressures is (A) $\sqrt{2}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{3}{\sqrt{2}}$ (D) $\sqrt{2}/3$

    ›Reveal solutionSolution

    This tests comparing final pressures reached by the same starting gas volume under two different doubling processes — adiabatic vs isothermal — using their respective $PV$ relations.

    Concept and Intuition

    An adiabatic expansion follows $PV^\gamma=\text{constant}$ while an isothermal expansion follows the simpler $PV=\text{constant}$ (Boyle's law). Both samples start with the same volume $V_0$ (though possibly different initial pressures) and both are expanded to the same final volume $2V_0$. Since we're told their final pressures come out equal, we can write each process's pressure relation, solve each for the common final pressure, and equate them to find the ratio of the two initial pressures needed to make that happen.

    Step-by-Step Solution

    1. Sample 1 undergoes adiabatic expansion: $P_1V_0^\gamma = P_f(2V_0)^\gamma \Rightarrow P_f = P_1\left(\dfrac{V_0}{2V_0}\right)^\gamma = \dfrac{P_1}{2^\gamma}$.
    2. Sample 2 undergoes isothermal expansion: $P_2V_0 = P_f(2V_0) \Rightarrow P_f=\dfrac{P_2}{2}$. …

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11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.A certain volume of a gas at 300 K expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas $= 1.5$) (A) $88\ \mathrm{K}$ (B) $77\ \mathrm{K}$ (C) $67\ \mathrm{K}$ (D) $54\ \mathrm{K}$

    ›Reveal solutionSolution

    Using the adiabatic relation $TV^{\gamma-1}=\text{const}$ with $\gamma=1.5$ and volume doubling gives a temperature fall of about $88\ \text{K}$.

    Concept and Intuition

    In an adiabatic process no heat is exchanged, so any expansion is powered entirely by the gas's own internal energy — the gas does work on its surroundings and necessarily cools. The relation $TV^{\gamma-1}=\text{constant}$ captures exactly how much a given volume expansion cools an ideal gas, with the exponent set by $\gamma$ (how many degrees of freedom store the energy).

    Step-by-Step Solution

    1. Adiabatic condition: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
    2. Rearranged: $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = 300\left(\dfrac{1}{2}\right)^{1.5-1} = 300(0.5)^{0.5}$.
    3. $(0.5)^{0.5} = 0.7071$, so $T_2 = 300\times0.7071 = 212.1\ \text{K}$. …

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12. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.An ideal gas is found to obey $PV^{3/2} = \text{constant}$ during an adiabatic process. If such a gas initially at a temperature T is adiabatically compressed to $\dfrac{1}{4}^{th}$ of its volume, then its final temperature is (A) $\sqrt{3}\,T$ (B) $\sqrt{2}\,T$ (C) $2T$ (D) $3T$

    ›Reveal solutionSolution

    Matching $PV^{3/2}=$const to the adiabatic form $PV^\gamma=$const gives $\gamma=3/2$; the companion relation $TV^{\gamma-1}=$const then gives the final temperature, $2T$.

    Concept and Intuition

    An adiabatic process for an ideal gas obeys $PV^\gamma=$const, and combining this with the ideal gas law $PV=nRT$ gives an equivalent relation directly between temperature and volume, $TV^{\gamma-1}=$const. Once we identify the effective $\gamma$ from the given $P$–$V$ relation, we can jump straight to how temperature responds to a volume change, without needing pressure at all.

    Step-by-Step Solution

    1. Given $PV^{3/2}=$const, so $\gamma=3/2$ for this (unusual) gas.
    2. Adiabatic $T$–$V$ relation: $TV^{\gamma-1}=$const, here $\gamma-1=1/2$. …

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