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