A) 857 J
B) 757 J
C) 1057 J
D) 957 J
Correct Answer: D
Solution :
For adiabatic change \[p{{V}^{\gamma }}=\] constant \[\therefore \] \[{{p}_{1}}{{V}_{1}}^{\gamma }={{p}_{2}}{{V}_{2}}^{\gamma }\] Also, \[{{C}_{V}}=\frac{3}{2}R\] From Mayors formula \[{{C}_{p}}-{{C}_{V}}=R\] \[\therefore \] \[{{C}_{p}}=R+{{C}_{V}}=R+\frac{3}{2}R=\frac{5}{2}R\] \[\gamma =\frac{{{C}_{p}}}{{{C}_{V}}}=\frac{\frac{5}{2}R}{\frac{3}{2}}=\frac{5}{3}\] \[\therefore \] \[{{p}_{2}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}{{p}_{1}}\] \[={{\left( \frac{6}{2} \right)}^{5/3}}\times {{10}^{5}}={{3}^{5/3}}\times {{10}^{5}}\] \[\Rightarrow \] \[{{p}_{2}}=6.19\times {{10}^{5}}N/{{m}^{2}}\] Work done during adiabatic expansion of gas is \[W=\frac{1}{1-\gamma }({{p}_{2}}{{V}_{2}}-{{p}_{1}}{{V}_{1}})\] \[\Rightarrow \] \[W=\frac{1}{1-\frac{5}{3}}\] \[\times 6.19\times {{10}^{5}}\times 2\times {{10}^{-3}}-{{10}^{5}}\times 6\times {{10}^{-3}})\] \[\Rightarrow \] \[W=\frac{3\times {{10}^{2}}\times 2}{2}(6.19-3)\] = 957 J Hence, external work done on gas will be 957 J.
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