Gujarati
Hindi
11.Thermodynamics
normal

$5.6$ liter of helium gas at $STP$ is adiabatically compressed to $0.7$ liter. Taking the initial temperature to be $T_1$, the work done in the process is

A

$\frac{9}{8}\,R{T_1}$

B

$\frac{3}{2}\,R{T_1}$

C

$\frac{15}{8}\,R{T_1}$

D

$\frac{9}{2}\,R{T_1}$

Solution

$\mathrm{WD}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}-\mathrm{P}_{2} \mathrm{V}_{2}}{\gamma-1}=\frac{\mathrm{nR}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1}$

$\mathrm{T}_{1}(5.6)^{\gamma-1}=\mathrm{T}_{2}(\mathrm{O} .7)^{\gamma-1} ; \mathrm{n}=\frac{5.6}{22.4}=\frac{1}{4}$

values $\mathrm{WD}=\frac{9}{8} \mathrm{RT}_{1}$

Standard 11
Physics

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