A) \[\frac{5}{4}R\]
B) \[\frac{31}{30}R\]
C) \[\frac{27}{32}R\]
D) \[\frac{30}{25}R\]
Correct Answer: B
Solution :
| In first process, |
| \[\Delta Q=\Delta U+0\]\[\Rightarrow \]\[{{C}_{V}}\Delta T=\frac{R}{\gamma -1}\cdot \Delta T\] |
| Also, \[pV=RT\]and \[3pV=R\,(T+\Delta T)\] |
| \[\Rightarrow \]\[2pV=R\Delta T\] |
| \[\therefore \]\[2RT=R\Delta T\]\[\Rightarrow \]\[\Delta T=2T\] |
| Hence, final temperature is 3T. |
| In second process, |
| \[(3p)V=R(3T)\] |
| \[\therefore \] \[{{T}_{\text{Final}}}=\frac{6RT}{R}=6T\] |
| and \[\Delta Q={{C}_{V}}(6T-3T)+3\phi \,(2V-V)\] |
| \[\Rightarrow \]\[\Delta Q={{C}_{V}}3T+{{3}_{P}}V\] |
| \[=\frac{R\,(3T)}{\gamma -1}+3RT\] |
| So, total heat given is |
| \[{{Q}_{\text{Total}}}=\frac{R(2T)}{\gamma -1}+\frac{R(3T)}{\gamma -1}+3RT\] |
| \[=\frac{5RT}{\gamma -1}+3RT\] |
| As, \[{{Q}_{\text{Total}}}=C\,(6T-T)\] |
| We have, |
| \[5C=\frac{5R}{\gamma -1}+3R\] |
| \[=\frac{R}{7/5-1}+\frac{3}{5}R=\frac{31}{30}R\] |
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