A) 28 K
B) 20 K
C) 15 K
D) 10 K
Correct Answer: A
Solution :
For cylinder A, pressure remains constant. Heat supplied \[Q=n{{C}_{p}}\Delta T\] Here, \[\Delta R=20\,\,K\] \[\therefore \] \[Q=n{{C}_{p}}\times 20\] ... (i) For cylinder B, volume remains constant. Here, heat supplied \[Q=n{{C}_{V}}\Delta T\]... (ii) From Eqs. (i) and (ii), we get \[n{{C}_{p}}\times 20=n{{C}_{V}}\Delta T\] For diatomic gas, \[{{C}_{p}}=\frac{7}{2}R\] and \[{{C}_{V}}=\frac{5}{2}R\] \[\frac{7}{2}R\times 20=\frac{5}{2}R\,\Delta T\,\,\,\,\,\Rightarrow \,\,\,\,\Delta T=28\,\,K\]
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