5.6,liter of helium gas at STP is adiabatically compressed to 0.7,liter . Taking initial temperature

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11.Thermodynamics

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$5.6\,liter$ 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

$\mathrm{T}_{1} \mathrm{V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{V}_{2}^{\gamma-1}$

$300 \times 5.6^{5 / 3-1}=\mathrm{T}_{2} \times(0.7)^{5 / 3-1}$

$W=\frac{n R\left(T_{1}-T_{2}\right)}{\gamma-1}$

Standard 11

Physics

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