A) 10 R
B) 50 R
C) 25 R
D) \[\left( \frac{25}{2} \right)R\]
Correct Answer: C
Solution :
\[\Delta Q=n{{C}_{v}}\Delta T\] \[\therefore \] \[25R=(1){{C}_{v}}(310-300)\]or \[{{C}_{v}}=\frac{5}{2}R\] i.e., the gas is diatomic or \[\gamma =1.4\] Now, work done in adiabatic process \[W=\frac{nR({{T}_{i}}+{{T}_{f}})}{\gamma -1}\] \[=\frac{1.R(310-300)}{1.4-1}=25R\] Hence, the correction option is [c].You need to login to perform this action.
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