The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the differe

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Q. The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

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Concept understanding — Heat Capacity at Constant Pressure

Heat Capacity at Constant Pressure — From Intuition to Precision

Imagine you have a pot of water on a stove. You turn the burner on, and the water gets hotter. How much heat does it take to raise its temperature by, say, 10°C? That depends on two things: how much water you have, and whether the pot is open to the air or sealed tight.

If the pot is open (constant pressure — the air above it is always at atmospheric pressure), the water can expand as it heats. Some of the energy you supply goes into pushing the atmosphere aside — doing work against the outside air. So you need to put in more heat than if the pot were sealed (constant volume), where no expansion work is possible.

That extra heat is the key idea behind heat capacity at constant pressure, denoted $C_p$ .

The Intuition First

Heat capacity tells you: "How much heat must I add to raise the temperature of this substance by 1°C (or 1 K)?"

At constant volume ( $C_v$ ): All the heat goes into increasing the internal energy (the kinetic and potential energy of the molecules). No work is done because the volume doesn't change.
At constant pressure ( $C_p$ ): Some heat goes into internal energy, but some also goes into the work of expansion against the constant external pressure. So $C_p$ is always larger than $C_v$ for gases (and for most solids/liquids, the difference is tiny because they barely expand).

Note

For an ideal gas, the difference is exactly $C_p - C_v = nR$ , where $n$ is the number of moles and $R$ is the universal gas constant. This is a direct consequence of the first law of thermodynamics.

The Precise Statement

Heat capacity at constant pressure is defined as the amount of heat required to raise the temperature of a substance by 1 K (or 1°C) while keeping the pressure constant.

Mathematically:

C_p = \left( \frac{\delta Q}{dT} \right)_p

The subscript $p$ means "at constant pressure." The $\delta Q$ (not $dQ$ ) reminds us that heat is a path-dependent quantity, not a state function.

But we can rewrite this in terms of a state function — enthalpy ( $H$ ). At constant pressure, the heat added equals the change in enthalpy:

\delta Q_p = dH

Therefore:

C_p = \left( \frac{\partial H}{\partial T} \right)_p

This is the working definition you'll use in problems: $C_p$ is the partial derivative of enthalpy with respect to temperature at constant pressure.

Molar vs. Specific Heat Capacity

You'll encounter two common forms:

Molar heat capacity at constant pressure ( $C_{p,m}$ ): heat capacity per mole (units: J mol⁻¹ K⁻¹)
Specific heat capacity at constant pressure ( $c_p$ ): heat capacity per unit mass (units: J kg⁻¹ K⁻¹)

The total heat capacity of a sample is:

C_p = n \cdot C_{p,m} = m \cdot c_p

Why It Matters

In most chemical reactions and physical processes, the system is open to the atmosphere — constant pressure. So $C_p$ is the relevant quantity for:

Calculating enthalpy changes ( $\Delta H = n C_{p,m} \Delta T$ )
Designing calorimeters (like coffee-cup calorimeters that operate at constant pressure) …

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The difference $C_P - C_V$ for an ideal gas is $nR$ , independent of the gas and the temperature. For 10 moles, this difference is $10R \approx 83.14\ \text{J K}^{-1}$ .

The relation $H = U + pV$ is the definition of enthalpy. For an ideal gas, $pV = nRT$ , so $H = U + nRT$ . The heat capacities at constant pressure and constant volume are defined as the partial derivatives of enthalpy and internal energy with respect to temperature:

C_P = \left(\frac{\partial H}{\partial T}\right)_p, \quad C_V = \left(\frac{\partial U}{\partial T}\right)_V

The key insight is that for an ideal gas, internal energy $U$ depends only on temperature, not on volume or pressure. This means $\left(\frac{\partial U}{\partial T}\right)_V = \frac{dU}{dT}$ , the same derivative regardless of the constraint. Similarly, enthalpy $H = U + nRT$ also depends only on temperature for an ideal gas, so $\left(\frac{\partial H}{\partial T}\right)_p = \frac{dH}{dT}$ .

Start from the definition of enthalpy: $H = U + pV$ . For an ideal gas, $pV = nRT$ , so $H = U + nRT$ .
Differentiate $H$ with respect to temperature at constant pressure:

C_P = \left(\frac{\partial H}{\partial T}\right)_p = \frac{dU}{dT} + nR

The constant-volume heat capacity is:

C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{dU}{dT}

Subtract the two expressions: $C_P - C_V = \left(\frac{dU}{dT} + nR\right) - \frac{dU}{dT} = nR$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

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

Cylinder P has a movable piston $\Rightarrow$ constant-pressure process: $Q = nC_p\Delta T_P$ .

Cylinder Q has a fixed piston $\Rightarrow$ constant-volume process: $Q = nC_v\Delta T_Q$ .

Both cylinders receive the same heat $Q$ and contain the same amount ( $n$ ) of the same diatomic gas, so: $nC_p\Delta T_P = nC_v\Delta T_Q \Rightarrow \Delta T_Q = \frac{C_p}{C_v}\Delta T_P.$ …
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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

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 $C_p = \frac{7}{2}R$ .

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 $\Delta T_1 = 0.42\, T_1$ .

Find $Q$ in terms of $T_1$

Substitute $\Delta T_1$ :

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

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

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:

$\frac{\Delta V_2}{V_2} = \frac{0.3 T}{T} = 0.3 = 30\%.$ …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\dfrac{2}{5}$
›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

At constant pressure, heat supplied: $Q = nC_p\Delta T$ . …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$^{-1}$
›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

$W = nR\Delta T$ (constant pressure); $\Delta U = nC_v\Delta T$ .

Given $W = 0.4\,\Delta U$ : $nR\Delta T = 0.4\times nC_v\Delta T$ .

The $n\Delta T$ cancels: $R = 0.4\,C_v \Rightarrow C_v = \dfrac{R}{0.4} = 2.5R$ . …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$= \dfrac{4}{3}$
›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

Fraction of heat converted to work at constant pressure: $\dfrac{W}{Q} = \dfrac{R}{C_p} = \dfrac{\gamma-1}{\gamma}$ . …
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AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$T \propto \dfrac{1}{\sqrt{V}}$
›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

Given: $T \propto V^{-1/2}$ , i.e. $T V^{1/2} = \text{const}$ .

Compare exponents with the adiabatic relation $T V^{\gamma - 1} = \text{const}$ .

Matching powers of $V$ : $\gamma - 1 = \dfrac{1}{2}$ . …
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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

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

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

Take the given ratio:

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

With $M_{H_2}=2\text{ g/mol}$ : …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$8.3\ \mathrm{J\,mol^{-1}K^{-1}}$
›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

Moles of oxygen: $n = \dfrac{20\ \text{g}}{32\ \text{g/mol}} = 0.625$ mol.

Temperature change: $\Delta T = 59 - 27 = 32$ K.

Molar heat capacity at constant pressure (diatomic): $C_p = \dfrac72 R = 3.5 \times 8.3 = 29.05\ \mathrm{J\,mol^{-1}K^{-1}}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\dfrac{16R}{7}$
›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

$H_2$ is diatomic: $C_{p,H_2} = \dfrac72R$ . Moles $n_1 = 4$ .

He is monatomic: $C_{p,He} = \dfrac52R$ . Moles $n_2 = 2$ .

Mixture's molar $C_p$ : …
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AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$\dfrac{4}{3}$
›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

First law: $\Delta U = Q - W = 80 - 20 = 60$ J.

$Q = nC_p\Delta T = 80$ J and $\Delta U = nC_v\Delta T = 60$ J. …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\gamma$
›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

Internal energy change: $\Delta U = nC_v\Delta T$ .

At constant pressure, $nR\Delta T = P\Delta V$ , so $nC_v\Delta T = \dfrac{C_v}{R}(P\Delta V)$ .

Since $C_v = \dfrac{R}{\gamma-1}$ , we get $\Delta U = \dfrac{P\Delta V}{\gamma - 1}$ .

Given $V \to 3V$ at constant $P$ : $\Delta V = 3V - V = 2V$ . …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$4/5$
›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

For a diatomic ideal gas (5 degrees of freedom: 3 translational + 2 rotational), $C_v=\tfrac52R$ and $C_p=C_v+R=\tfrac72R$ . …
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Q.The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

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