A) 5
B) 6.5
C) 7
D) 3.5
Correct Answer: A
Solution :
As gas is suddenly expanded so it is an adiabatic process, \[ie\], \[p{{V}^{\gamma }}=\]constant or \[{{p}_{1}}V_{1}^{\gamma }={{p}_{2}}V_{2}^{\gamma }\] Given \[{{V}_{2}}=3{{V}_{1}},\,\,{{C}_{V}}=2R\] \[\therefore \] \[{{C}_{p}}=2R+R=3R\] \[\Rightarrow \] \[\gamma =\frac{{{C}_{p}}}{{{C}_{V}}}=\frac{3R}{2R}=1.5\] \[\therefore \] \[\frac{{{p}_{1}}}{{{p}_{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma }}={{(3)}^{1.5}}=5.1\approx 5\]
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