For a gas $$C_P - C_V = R$$ in a state P and $$C_P - C_V = 1.10R$$ in a state Q. $$T_P$$ and $$T_Q$$ are the temperatures in two different states P and Q, respectively. Then

The difference between the molar heat capacities of any fluid is given by the exact thermodynamic identity

$$C_P-C_V \;=\; -\,T\; \frac{\left(\dfrac{\partial P}{\partial T}\right)_{V}^{2}} {\left(\dfrac{\partial P}{\partial V}\right)_{T}}.$$

For an ideal gas the equation of state is $$P\,V = R\,T$$ and after inserting this equation in the above identity one gets the familiar Mayer relation $$C_P-C_V=R,$$ which is independent of temperature. A departure of the value of $$C_P-C_V$$ from $$R$$ therefore signals non-ideal behaviour, so we must work with a more realistic equation of state. The simplest and most frequently used one is the van der Waals equation

$$P=\frac{R\,T}{V-b}-\frac{a}{V^{2}},$$

in which the parameters $$a$$ and $$b$$ account, respectively, for intermolecular attraction and the finite size of the molecules.

We now calculate the two derivatives that enter the heat-capacity identity.

First, keeping volume constant,

$$\left(\frac{\partial P}{\partial T}\right)_{V} =\frac{R}{V-b}.$$

Secondly, keeping temperature constant,

$$\left(\frac{\partial P}{\partial V}\right)_{T} = -\frac{R\,T}{(V-b)^{2}}+\frac{2a}{V^{3}}.$$

Substituting these derivatives into the general expression, we obtain

$$C_P-C_V = -\,T\; \frac{\displaystyle\left(\dfrac{R}{V-b}\right)^{2}} {\displaystyle -\frac{R\,T}{(V-b)^{2}}+\frac{2a}{V^{3}}}.$$

The two negative signs in the numerator and denominator cancel, and a factor $$\dfrac{R\,T}{(V-b)^{2}}$$ can be taken out of the denominator to give

$$C_P-C_V = \frac{T\,R^{2}/(V-b)^{2}} {\,R\,T/(V-b)^{2}\, \Bigl[\,1-\dfrac{2a(V-b)^{2}}{R\,T\,V^{3}}\Bigr]} = \frac{R}{1-\dfrac{2a(V-b)^{2}}{R\,T\,V^{3}}}.$$

Write this result in the more compact form

$$C_P-C_V =\;R\; \Biggl[\, 1-\frac{2a\,(V-b)^{2}}{R\,T\,V^{3}} \Biggr]^{-1}. \quad -(1)$$

Observe the structure of the denominator in (1). The fraction

$$\frac{2a\,(V-b)^{2}}{R\,T\,V^{3}}$$

is strictly positive (all the symbols denote positive quantities) and it is divided by $$T$$. Hence

• When the temperature $$T$$ is very high, the term $$\dfrac{2a\,(V-b)^{2}}{R\,T\,V^{3}}$$ becomes small, the bracket approaches $$1$$ and (1) reduces to $$C_P-C_V\approx R$$ - the ideal-gas limit.
• When the temperature $$T$$ is lowered, the same term becomes larger, the entire bracket becomes smaller than 1, and therefore its reciprocal, and with it $$C_P-C_V,$$ becomes larger than $$R$$.

In other words, the difference $$C_P-C_V$$ is inversely related to the temperature for a real gas described by the van der Waals equation: a greater value of $$C_P-C_V$$ corresponds to a lower temperature.

Let us now compare the two given states.

State $$P$$: $$C_P-C_V = R$$ (the ideal value)
State $$Q$$: $$C_P-C_V = 1.10\,R$$ (10 % larger than the ideal value)

Because $$1.10\,R \;>\; R,$$ the above reasoning tells us that state $$Q$$ must be at a lower temperature than state $$P$$. Symbolically,

$$T_Q \;< T_P.$$

So, the correct inequality is

$$T_P \;>\; T_Q.$$

Hence, the correct answer is Option D.