A) 21 K
B) 35 K
C) 42 K
D) 70 K
Correct Answer: C
Solution :
From first law of thermodynamics \[Q=\Delta U+W\] For cylinder A pressure remains constant \[\therefore \] Work done by a system \[W=\frac{\mu R}{\gamma -1}({{T}_{1}}-{{T}_{2}})\] For monoatomic gases \[\mu =\frac{1}{5}\] \[\gamma =\frac{5}{3}\] \[\therefore \] \[W=\frac{\frac{1}{5}\times R}{\frac{5}{3}-1}(442-400)=\frac{3}{2}R\times 42\] or \[W=63R\] But \[\Delta U=0,\] for cylinder A \[\therefore \] \[Q=0+63R\] \[Q=63R\] For cylinder B volume is constant, \[\therefore \] \[W=0\] 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 to both cylinder is same \[\therefore \] \[63R=\frac{3}{2}R\Delta T\] \[\Delta T=42K\]
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