Analysis
We are given a mono-atomic gas (Degrees of freedom $f=3$, $C_V = \frac{3R}{2}$) undergoing a process where the molar specific heat is $C = 2R$.
The general formula for molar heat capacity in a polytropic process $PV^x = \text{constant}$ is given by:
$$ C = C_V + \frac{R}{1-x} $$Substituting the given values:
$$ 2R = \frac{3R}{2} + \frac{R}{1-x} $$ $$ \frac{R}{2} = \frac{R}{1-x} \implies 1-x = 2 \implies x = -1 $$So the process follows the equation $PV^{-1} = \text{constant}$, which implies $P \propto V$.
Temperature-Volume Relationship
Using the ideal gas equation $PV = nRT$, we can substitute $P \propto \frac{T}{V}$ into the process equation:
$$ \left(\frac{T}{V}\right) V^{-1} = \text{constant} \quad \text{(Incorrect substitution)} $$Let’s use the standard relation for polytropic process $TV^{x-1} = \text{constant}$:
$$ T V^{-1-1} = \text{constant} \implies T V^{-2} = \text{constant} $$ $$ T \propto V^2 $$Alternatively, since $P \propto V$, substituting into $PV \propto T$ gives $(V)V \propto T \implies V^2 \propto T$.
Calculation
The volume is doubled: $V_f = 2V_i$.
$$ \frac{T_f}{T_i} = \left( \frac{V_f}{V_i} \right)^2 = (2)^2 = 4 $$Thus, the temperature becomes 4 times the initial temperature.
Correct Option: (d) 4