A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in the process is given, (given γ = 3 2 ) :

Question

A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in the process is given, (given &#947; = 3 2 ) :

Aicharya · Accepted Answer

Work done by gas in adiabatic process is:

$W = - \frac{(P_{2} V_{2} - P_{1} V_{1})}{γ - 1}$

$= - \frac{n R (T_{2} - T_{1})}{γ - 1}$

For an adiabatic process, $T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1}$

$\Rightarrow T V^{\frac{3}{2} - 1} = T_{2} {(2 V)}^{\frac{3}{2} - 1} \Rightarrow T V^{\frac{1}{2}} = T_{2} {(2 V)}^{\frac{1}{2}} \Rightarrow T_{2} = \frac{T}{\sqrt{2}}$

Now, work done is $W = \frac{R (T_{1} - T_{2})}{γ - 1} = \frac{R (T - \frac{T}{\sqrt{2}})}{\frac{1}{2}} = R T (2 - \sqrt{2})$

A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in…

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