GATE Overflow for GATE XE - Recent questions and answers in Thermodynamic Relations

GATE XE 2026 | Question: 87

Tue, 24 Feb 2026 15:40:19 +0000

Consider a real gas, which obeys the following equation-of-state:

$$v=\dfrac{R T}{P}+C_{1}-\dfrac{C_{2}}{P v},$$

where $v$ is the mass specific volume, $P$ is the pressure, $T$ is the temperature and $R$ is the gas constant. $C_{1}$ and $C_{2}$ are constants. If $s$ is the mass specific entropy, the quantity $\left(\dfrac{\partial s}{\partial v}\right)_{T}$ for the gas is given by

$\dfrac{R}{v}$
$\dfrac{R^{2} T}{P v^{2}}$
$\dfrac{R}{v-C_{1}}$
$\dfrac{C_{1} C_{2}}{T v}$

GATE XE 2026 | Question: 89

Tue, 24 Feb 2026 15:39:56 +0000

A fluid undergoes a process where its pressure $(P)$, temperature $(T)$ and volume $(V)$ changes from $\left(P_{1}, T_{1}, V_{1}\right)$ to $\left(P_{2}, T_{2}, V_{2}\right)$. During the process, volume expansivity ( $\beta$ ) and isothermal compressibility ( $\kappa_{T}$ ) remains constant. Given that $\beta=\dfrac{1}{V}\left(\dfrac{\partial V}{\partial T}\right)_{P}$ and $\kappa_{T}=-\dfrac{1}{V}\left(\dfrac{\partial V}{\partial P}\right)_{T}$, the ratio $\left(\dfrac{V_{2}}{V_{1}}\right)$ is

$\dfrac{\beta\left(T_{2}-T_{1}\right)}{\kappa_{T}\left(P_{2}-P_{1}\right)}$
$\left[\beta\left(T_{2}-T_{1}\right)\right]\left[\kappa_{T}\left(P_{2}-P_{1}\right)\right]$
$\dfrac{\exp \left[\beta\left(T_{2}-T_{1}\right)\right]}{\exp \left[\kappa_{T}\left(P_{2}-P_{1}\right)\right]}$
$\beta\left(T_{2}-T_{1}\right) - \kappa_{T}\left(P_{2}-P_{1}\right)$

GATE XE 2026 | Question: 94

Tue, 24 Feb 2026 15:38:56 +0000

It is given that: $\left(\dfrac{\partial h}{\partial T}\right)_{P}=c_{p}$ and $\left(\dfrac{\partial h}{\partial P}\right)_{T}=(v-\beta T v)$. Here $h$ is the mass specific enthalpy, $v$ is the mass specific volume, $\beta=\dfrac{1}{v}\left(\dfrac{\partial v}{\partial T}\right)_{P}$ is the volume expansivity, and $c_{p}$ is the mass specific heat capacity at constant pressure. $T$ and $P$ represent the temperature and the pressure, respectively. The inversion temperature is the temperature at which the Joule-Thomson coefficient, $\mu_{\mathrm{JT}}=\left(\frac{\partial T}{\partial P}\right)_{h}$, goes to zero. Consider a fluid with properties: $v=1.03 \mathrm{~m}^{3} / \mathrm{kg}, c_{P}=1 \mathrm{~kJ} / \mathrm{kg}-\mathrm{K}$, and $\beta=4.39 \times 10^{-3} 1 / \mathrm{K}$;

The inversion temperature (in $\mathrm{K}$) for the fluid is $\_\_\_\_$ (rounded off to two decimal places).

GATE XE 2025 | Question: 62

Sun, 04 May 2025 19:06:58 +0000

The standard free energy change for the reaction, $\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{SO}_{3}$ at equilibrium is given by $\Delta \text{G}^{\circ}=-94600+89.37 \mathrm{~T}$, where $\text{T}$ is in Kelvin and $\Delta \mathrm{G}^{\circ}$ is in Joule. The equilibrium constant $\left(\mathrm{K}_{\mathrm{P}}\right)$ at $1050 \: \mathrm{K}$ is (rounded off to two decimal places) $\_\_\_\_\_\_\_\_$

Given: Universal gas constant $(\mathrm{R})=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

GATE XE 2024 | Question: 79

Sun, 21 Jul 2024 16:41:40 +0000

Given $v$ is the molar specific volume, $P$ is the pressure, $T$ is the temperature, $R$ is the Universal gas constant, and $a, b$ are van der Waal's constants.
The van der Waal's equation of state is
\[
P=\frac{R T}{v-b}-\frac{a}{v^{2}}
\]

The value of $\left(\frac{\partial v}{\partial T}\right)_{P}\left(\frac{\partial P}{\partial v}\right)_{T}\left(\frac{\partial T}{\partial P}\right)_{v}$ is

$\frac{a}{b^{2}}$
$-1$
$1$
$\frac{b^{2}}{a}$

GATE XE 2024 | Question: 88

Sun, 21 Jul 2024 16:41:32 +0000

In a graph with Helmholtz function on the $y$-axis and volume on the $x$-axis, the slope of the isothermal curves for a finite volume system containing an ideal gas is

always zero
infinite
finite, positive, and non-zero
finite, negative, and non-zero

GATE XE 2024 | Question: 92

Sun, 21 Jul 2024 16:41:28 +0000

The equation of state for a non-ideal gas is
\[
\frac{P v}{R T}=1+B P
\]
where $P$ is pressure, $v$ is specific volume, $R$ is the specific gas constant, $T$ is temperature, and $B$ is a temperature dependent parameter. For this gas, the partial derivative of enthalpy with respect to pressure at constant temperature is

$B R T$
$-R T^{2}\left(\frac{d B}{d T}\right)$
$B R T-R T^{2}\left(\frac{d B}{d T}\right)$
$0$