A) 4 arm
B) \[\frac{3}{2}\]arm
C) 8 aim
D) \[\frac{1}{4}\]arm
Correct Answer: C
Solution :
The adiabatic relation between volume and pressure of a gas \[P{{V}^{\gamma }}=\] constant or \[{{P}_{2}}{{V}_{2}}^{\gamma }={{P}_{1}}{{V}_{1}}^{\gamma }\] or \[{{P}_{2}}={{P}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}\] ... (i) Given, \[{{P}_{1}}=1\,atm\,,\,{{V}_{1}}=V,\,\,{{V}_{2}}=\frac{V}{4},\,\,\gamma =\frac{3}{2}\] Substituting in Eq. (i), we get \[{{P}_{2}}=1\times {{\left( \frac{V}{V/4} \right)}^{3/2}}={{(4)}^{3/2}}=8\,atm\]
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