The specific heat at constant pressure of a real gas obeying P V^2=R T equation is:

Heat and Thermodynamics PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q62 with Solution

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Correct answer: C

Use the general relation for any gas

For any thermodynamic system,

C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P

We are given the equation of state of the real gas:

PV^2 = RT

So,

P = \frac{RT}{V^2}

Compute $\left(\frac{\partial P}{\partial T}\right)_V$

At constant $V$ ,

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V^2}

Compute $\left(\frac{\partial V}{\partial T}\right)_P$

From

PV^2 = RT

at constant $P$ ,

V^2 = \frac{RT}{P}

Differentiate with respect to $T$ at constant $P$ :

2V\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{P}

Thus,

\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{2PV}

Now use $P=\dfrac{RT}{V^2}$ :

\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{2\left(\frac{RT}{V^2}\right)V} = \frac{V}{2T}

Substitute into the formula

C_P - C_V = T\left(\frac{R}{V^2}\right)\left(\frac{V}{2T}\right)

So,

C_P - C_V = \frac{R}{2V}

Hence,

C_P = C_V + \frac{R}{2V}

Match with the options

This corresponds to:

\boxed{C_P = C_V + \frac{R}{2V}}

So the correct option is C.

Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: C

Use the general relation for any gas

For any thermodynamic system,

C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P

We are given the equation of state of the real gas:

PV^2 = RT

So,

P = \frac{RT}{V^2}

Compute $\left(\frac{\partial P}{\partial T}\right)_V$

At constant $V$ ,

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V^2}

Compute $\left(\frac{\partial V}{\partial T}\right)_P$

From

PV^2 = RT

at constant $P$ ,

V^2 = \frac{RT}{P}

Differentiate with respect to $T$ at constant $P$ :

2V\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{P}

Thus,

\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{2PV}

Now use $P=\dfrac{RT}{V^2}$ :

\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{2\left(\frac{RT}{V^2}\right)V} = \frac{V}{2T}

Substitute into the formula

C_P - C_V = T\left(\frac{R}{V^2}\right)\left(\frac{V}{2T}\right)

So,

C_P - C_V = \frac{R}{2V}

Hence,

C_P = C_V + \frac{R}{2V}

Match with the options

This corresponds to:

\boxed{C_P = C_V + \frac{R}{2V}}

So the correct option is C.

Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.

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Heat and Thermodynamics question

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