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\[=\frac{Q}{\Delta T}=\frac{\frac{19}{2}R{{T}_{0}}}{3{{T}_{0}}}=\frac{19}{6}R\]You need to login to perform this action.
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