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.

COMEDK 2026Set 2026-M1 markMCQ
$\frac{5 R}{2}$
›Reveal solutionSolution

Using $Q = nC_p\Delta T$ at constant pressure with $C_p = \tfrac{5R}{2}$ gives $n \approx 1.6$ moles.

Applying the first law at constant pressure

Heat supplied at constant pressure:
$Q = n\,C_p\,\Delta T$
Given $C_p = \dfrac{5R}{2}$ , $Q = 1662\,\text{J}$ and $\Delta T = 50\,\text{K}$ , with $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$ :

$C_p = \frac{5 \times 8.314}{2} = 20.785\,\text{J mol}^{-1}\text{K}^{-1}$ …
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KCET 2025Set D-41 markMCQ
$x$
›Reveal solutionSolution
Apply the steady-state junction rule — heat in = heat out — using $\dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L}$ ; with three identical rods the geometry factor cancels and $2(90 - x) = x$ gives $x = 60^\circ$ C.

Step 1 — The set-up (from the figure)

Three identical rods meet at a common junction $x$ :

One rod runs to an end held at $\mathbf{0^\circ C}$ (the cold sink).

Two rods run to ends held at $\mathbf{90^\circ C}$ (the hot sources).

The rods are of the same material and identical in all respects, so each has the same thermal conductivity $k$ , the same cross-sectional area $A$ and the same length $L$ . No heat escapes through the curved surfaces, so all the heat that arrives at the junction must leave through the rods.

Step 2 — The concept: steady-state conduction

The rate of heat conduction along a rod (Fourier's law) is
$\frac{dQ}{dt} = \frac{k A \,(T_{\text{hot}} - T_{\text{cold}})}{L}$
Define the thermal conductance $C = \dfrac{kA}{L}$ , which is identical for all three rods. Then simply
$\frac{dQ}{dt} = C\,\Delta T$
In the steady state the junction's temperature is no longer changing, which means it is storing no heat. Therefore
$\boxed{\text{(heat flowing IN per second)} = \text{(heat flowing OUT per second)}}$
This is the thermal analogue of Kirchhoff's junction rule for currents.

Step 3 — Write the heat currents

The junction is at temperature $x$ , with $0 < x < 90$ (it must lie between the extremes).

Heat IN — along the two rods from the $90^\circ$ C ends (heat flows hot → cold, so into the junction):
$\left(\frac{dQ}{dt}\right)_{\text{in}} = 2 \times C\,(90 - x)$
Heat OUT — along the one rod to the $0^\circ$ C end:
$\left(\frac{dQ}{dt}\right)_{\text{out}} = C\,(x - 0) = C\,x$
Step 4 — Equate and solve
$2\,C\,(90 - x) = C\,x$
The conductance $C = kA/L$ is the same on both sides, so it cancels entirely — which is why we never needed the values of $k$ , $A$ or $L$ :

$2(90 - x) = x$ …
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KCET 2024Set D-21 markMCQ
$4.5\,P_0V_0$
›Reveal solutionSolution
Net work in a cycle = area enclosed on the $P$ – $V$ diagram (positive for a clockwise loop); the loop is a right triangle of legs $2V_0$ and $2P_0$ , so $W = \tfrac12(2V_0)(2P_0) = 2P_0V_0$ .

Step 1 — The concept.

Work done by a gas is $W = \int P\,dV$ , which is the area under the path on a $P$ – $V$ diagram. Around a closed cycle the areas under the outgoing and returning legs partially cancel, leaving
$W_{\text{net}} = \oint P\,dV = \pm\,(\text{area enclosed by the loop}),$
positive if the cycle runs clockwise (the gas expands at high pressure and is compressed at low pressure — a net output, as in an engine) and negative if anticlockwise. Note also that for a full cycle $\Delta U = 0$ (internal energy is a state function), so by the first law $Q = W$ — but we do not need that here.

Step 2 — Read the vertices.
$M = (V_0,\ 3P_0),\qquad N = (3V_0,\ P_0),\qquad O = (V_0,\ P_0).$
The three legs are: M→N a straight sloping line, N→O horizontal at $P = P_0$ (compression), O→M vertical at $V = V_0$ (isochoric pressure rise).

Step 3 — Compute the enclosed area.

The triangle is right-angled at $O$ :

horizontal leg $ON = 3V_0 - V_0 = 2V_0$ ,

vertical leg $OM = 3P_0 - P_0 = 2P_0$ .

$\text{Area} = \frac12 \times \text{base} \times \text{height} = \frac12 (2V_0)(2P_0) = 2\,P_0V_0 .$
Step 4 — Fix the sign from the sense of traversal.

Apply the shoelace formula to $M(1,3) \to N(3,1) \to O(1,1)$ (in units of $V_0$ and $P_0$ ):
$\sum = x_M(y_N - y_O) + x_N(y_O - y_M) + x_O(y_M - y_N) = 1(1-1) + 3(1-3) + 1(3-1) = -6 + 2 = -4 .$
A negative shoelace sum means the vertices are listed clockwise, and $|{-4}|/2 = 2$ ⇒ area $= 2P_0V_0$ . Clockwise ⇒ the gas does positive net work. …
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KCET 2020Set A-11 markMCQ
$\frac{2}{3}$
›Reveal solutionSolution
For a monoatomic ideal gas expanding at constant pressure, the fraction of supplied heat converted into work is $\frac{2}{5}$ , which corresponds to option (C).

The key here is to connect the heat supplied at constant pressure to the two places it can go: increasing the internal energy of the gas, and doing work on the surroundings. The fraction that becomes work is simply the work done divided by the heat supplied. Since the process is at constant pressure, both the work and the heat have straightforward expressions in terms of the temperature change, and the ratio depends only on the gas's specific heat capacities — which for a monoatomic gas are fixed numbers.

Let’s walk through it step by step.

Identify the relevant thermodynamic quantities. For a monoatomic ideal gas, the molar specific heat at constant volume is $C_V = \frac{3}{2}R$ , and at constant pressure it is $C_P = C_V + R = \frac{5}{2}R$ . When heat $Q$ is supplied at constant pressure to $n$ moles, causing a temperature rise $\Delta T$ , we have:

$Q = n C_P \Delta T = n \left(\frac{5}{2}R\right) \Delta T.$

Find the work done during the expansion. At constant pressure $P$ , the work done by the gas is $W = P \Delta V$ . Using the ideal gas law $PV = nRT$ , a change in volume at constant pressure gives $P \Delta V = nR \Delta T$ . So:

$W = nR \Delta T.$

Compute the fraction of heat converted to work. The fraction is $W/Q$ : $\frac{W}{Q} = \frac{nR \Delta T}{n \cdot \frac{5}{2}R \Delta T} = \frac{1}{\frac{5}{2}} = \frac{2}{5}.$ …
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KCET 2019Set A-11 markMCQ
$O_2$
›Reveal solutionSolution
For an ideal diatomic gas heated at constant pressure, the heat required to double the volume is $Q = n C_p \Delta T$ . With $n=1$ , $C_p = \frac{7}{2}R$ , and $\Delta T = 300\,\text{K}$ (since $V \propto T$ at constant $P$ ), we get $Q = 1050\,R$ . The correct option is (D).

The key here is to recognise that the gas is oxygen ( $O_2$ ), which is diatomic. At constant pressure, the heat added equals the change in enthalpy: $Q = n C_p \Delta T$ . And because pressure is constant, volume is directly proportional to absolute temperature (Charles’s law). So doubling the volume means doubling the absolute temperature.

Let’s walk through it step by step.

Identify the gas and its degrees of freedom.

Oxygen ( $O_2$ ) is a diatomic molecule. At moderate temperatures (like $27^\circ\text{C}$ ), it has 5 degrees of freedom: 3 translational and 2 rotational. Vibrational modes are not excited at this temperature.

For a diatomic ideal gas:

$C_v = \frac{5}{2}R$

$C_p = C_v + R = \frac{7}{2}R$

Convert the initial temperature to Kelvin.

$T_1 = 27^\circ\text{C} = 300\,\text{K}$

Always use absolute temperature in gas law calculations.

Relate volume change to temperature change at constant pressure.

For an ideal gas at constant pressure: $V \propto T$ (Charles’s law).

So if volume doubles: $\frac{V_2}{V_1} = 2 \implies \frac{T_2}{T_1} = 2$

Therefore $T_2 = 2 \times 300 = 600\,\text{K}$ , and $\Delta T = 300\,\text{K}$ .

Heat added at constant pressure.

The heat required is $Q = n C_p \Delta T$ . …
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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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