A) \[\frac{5}{2}{{P}_{0}}\]
B) \[\frac{3}{2}{{P}_{0}}\]
C) \[2{{P}_{0}}\]
D) \[\frac{7}{2}{{P}_{0}}\]
Correct Answer: A
Solution :
Process AB pressure = constant \[V\propto T\] \[{{V}_{B}}=2{{V}_{0}}\] So, \[{{T}_{B}}=2{{T}_{0}}\] (If\[{{T}_{A}}={{T}_{0}}\]) \[\Delta W={{P}_{0}}(2{{V}_{0}}-{{V}_{0}})={{P}_{0}}{{V}_{0}}=R{{T}_{0}}\] \[\Delta Q={{C}_{V}}\Delta T=\frac{5}{2}R{{T}_{0}}\] \[\Delta U=\Delta Q=\Delta Q=\frac{3}{2}R{{T}_{0}}\] Process BC volume = constant, \[P\propto T\] \[{{P}_{B}}={{P}_{0}}\] but \[{{T}_{B}}=2{{T}_{0}}\] So, \[{{P}_{C}}=K{{P}_{0}},\,\,{{T}_{C}}=K{{T}_{B}}=K\times 2{{T}_{0}}\] \[\Delta W=0\] \[\Delta Q=\Delta U={{C}_{V}}\Delta T\] \[=\frac{3}{2}R\times 2{{T}_{0}}(K-1)\] \[=2R{{T}_{0}}(K-1)\] \[\Delta {{Q}_{total}}=\frac{5}{2}R{{T}_{0}}+3R{{T}_{0}}(K-1)\] ?(1) \[\Delta {{U}_{total}}=\frac{3}{2}R{{T}_{0}}+3R{{T}_{0}}(K-1)\] ?(2) \[\Delta U/\Delta Q=6/7\] Compare (1), (2) and (3) \[K=5/2\] So\[,\,\,P=\frac{5}{2}{{P}_{0}}\].
You need to login to perform this action.
You will be redirected in
3 sec
