The volume of $1 \; mole$ of an ideal gas with the adiabatic exponent $\gamma$ is changed according to the relation $V = \frac{b}{T}$,where $b$ is a constant. The amount of heat absorbed by the gas in the process if the temperature is increased by $\Delta T$ will be:

Consider a $P - V$ diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.
$(a)$ Find the work done when the gas is taken from state $1$ to state $2$.
$(b)$ What is the ratio of temperature $\frac{T_1}{T_2}$ if $V_2 = 2V_1$?
$(c)$ Given the internal energy for one mole of gas at temperature $T$ is $\frac{3}{2}RT$,find the heat supplied to the gas when it is taken from state $1$ to $2$,with $V_2 = 2V_1$.

The $P-V$ diagram of a diatomic gas is a straight line passing through the origin. The molar heat capacity of the gas in this process will be:

Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$, if it is expanding under the condition $V \propto T^{2/3}$ is, $(R = 8.314 \ J/mol \cdot K)$ (in $J$)

An ideal gas with heat capacity at constant volume $C_V$ undergoes a quasistatic process described by $p V^{\alpha} = \text{constant}$ in a $p-V$ diagram,where $\alpha$ is a constant. The heat capacity of the gas during this process is given by

$A$ gas expands in such a way that its pressure and volume satisfy the condition $PV^2 = \text{constant}$. Then the temperature of the gas