Question

If a monoatomic ideal gas of volume 1 litre at N.T.P I'd compressed (i) adiabatically to half of its volume, find
the work done
on
the gas. Also
find (ii) the work done
if the compression is isothermal (y =5/3)​

Answer 1

Answer:

1. $\rm{-88.5J}$
2. $\rm{-70J}$

Explanation:

During adiabatic process,

$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$

(i) $T_{1} = 273K, \:V_{1}=V ; \: V_{2} = \frac{V}{2}$

$T_{2} = 273\bigg(\frac{V}{V}\bigg)^{5/3-1}, \:\:\:T_{2} = 273\bigg(\frac{V}{V/2}\bigg)^{5/3-1} = 273\times 2^{2/3} = 431.6K$

Number of moles moles ($\mu$) = $\frac{1lit}{22.4lit}=\frac{1}{22.4}$

$W = \frac{\mu R}{\gamma - 1}(T_{1}-T_{2}),\:R = 8.314J \:mol^{-1}k^{-1},\:W = \frac{8.314}{22.4\bigg(\frac{5}{3}-1\bigg)}\times (274-431.6)$

$W = \frac{8.34\times3}{22.4\times2}(-158.6)=-88.5J$

(ii) Work done during isothermal compression is

$W = 2.3026\times \mu RTlog\bigg(\frac{V_{2}}{V_{1}}\bigg),\:\:\mu=\frac{1}{22.4},\:T= 273K,\:R = 8.314J\:mol^{-1}k^{-1}$

$\frac{V_{2}}{V_{1}}=\frac{1}{2}=0.5;$

$W = \frac{2.3026\times 8.314\times 273\times log_{10}(0.5)}{22.4}=-70J\:\:\:\:\:\bigg[\because\:\frac{V_{2}}{V_{1}}=\frac{V/2}{V}=\frac{1}{2}\bigg]$