System Analysis
Consider the entire cylinder contents (Gas 1 + Gas 2) as the system. The walls and piston are adiabatic (non-conducting), but the partition between them is conducting. This means:
- There is no heat exchange with the surroundings: $dQ_{total} = 0$.
- Both parts maintain the same temperature $T$ due to the conducting partition.
- Gas 2 is in a fixed volume (between fixed partition and closed end), so $dV_2 = 0$ and $dW_2 = 0$.
Heat Exchange
Since total heat exchange is zero: $dQ_1 + dQ_2 = 0 \implies dQ_1 = -dQ_2$.
For Gas 2, the process is isochoric (constant volume). Thus, heat absorbed is:
$$ dQ_2 = n_2 C_V dT $$Therefore, heat absorbed by Gas 1 is:
$$ dQ_1 = -n_2 C_V dT $$Molar Specific Heat of Gas 1
The molar specific heat $C$ is defined as heat absorbed per mole per unit temperature change:
$$ C_1 = \frac{dQ_1}{n_1 dT} = \frac{-n_2 C_V dT}{n_1 dT} = -\frac{n_2}{n_1} C_V $$For a mono-atomic gas, $C_V = \frac{3R}{2}$. Also, the ratio of moles is equal to the ratio of masses ($n = m/M$):
$$ C_1 = -\frac{m_2}{m_1} \left( \frac{3R}{2} \right) = -\frac{3m_2 R}{2m_1} $$Correct Option: (d) $-\frac{3m_2 R}{2m_1}$