5.6, liter of helium gas at STP is adiabatica | vedclass

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Question

Initially

$V_{1}=5.6 l, T_{1}=273 K, P_{1}=1 a t m$

$\gamma=\frac{5}{3}(F 	ext { or monoa } ightarrow 	ext { micgas })$

The number of mol

Vedclass · Accepted Answer

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

Vedclass · Answer

Initially

$V_{1}=5.6 l, T_{1}=273 K, P_{1}=1 a t m$

$\gamma=\frac{5}{3}(F 	ext { or monoa } ightarrow 	ext { micgas })$

The number of moles of gas is $n=\frac{5.6 l}{22.4 l}=\frac{1}{4}$

Finally (after adiabatic compression)

$V_{2}=0.7 l$

For adiabatic compression $T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$

$	herefore T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}ight)^{\gamma-1}=T_{1}\left(\frac{5.6}{0.7}ight)^{\frac{5}{3}-1}=T_{1}(8)^{2 / 3}=4 T_{1}$

We know that work done in adiabatic process is

$W=\frac{n R \Delta T}{\gamma-1}=\frac{9}{8} R T_{1}$

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

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