question_answer

A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until the volume is doubled. The average molar heat capacity for the whole process is

A) \[\frac{13R}{6}\]          

B) \[\frac{19R}{6}\]

C) \[\frac{23R}{6}\]          

D) \[\frac{17R}{6}\]

Correct Answer: B

Solution :

[b] Let initial pressure, volume and temperature be \[{{P}_{0}},{{V}_{0}}\]and\[{{T}_{0}}\], respectively, indicated by state A in P-V diagram. The gas is then isochorically taken to state B \[(2{{P}_{0}},{{V}_{0}},2{{T}_{0}})\]and then taken form state B to state C \[(2{{P}_{0}},2{{V}_{0}},4{{T}_{0}})\]isobarically. Total heat absorbed by 1 mol of gas \[Q={{C}_{v}}(2{{T}_{0}}-{{T}_{0}})+{{C}_{P}}(4{{T}_{0}}-2{{T}_{0}})\] \[=\frac{5}{2}R{{T}_{0}}+\frac{7}{2}R\times 2{{T}_{0}}=\frac{19}{2}R{{T}_{0}}\] Total change in temperature from series A to C is \[\Delta T=3{{T}_{0}}\] Therefore, Molar heat capacity\[=\frac{Q}{\Delta T}=\frac{\frac{19}{2}R{{T}_{0}}}{3{{T}_{0}}}=\frac{19}{6}R\]