VITEEE 2009 Paper

From first law of thermodynamics Q = ΔU + W For cylinder A pressure remains constant. ∴ Work done by a system $W=\frac{\mu R}{\gamma-1}(T_{1}-T_{2})$ For monoatomic gases, $\mu=1;\gamma=\frac{5}{3}$ ∴ $W=\frac{1\times R}{\frac{5}{3}-1}(442-400)=\frac{3}{2}R\times 42$ or W = 63 R But ΔU = 0, for cylinder A ∴ Q = 0 + 63R = 63R For cylinder B volume is constant, ∴ $W = 0 \text{ and } Q = \mu C_{v} \Delta T$ For monoatomic gas $C_{v}=\frac{3}{2}R$ ⇒ $Q = 1 \times \frac{3}{2} R \Delta T$ As heat given on both cylinder is same ∴ $63R = \frac{3}{2} R \Delta T$ ⇒ ΔT = 42 K