In a cylinder, $2.0 m o l e s$ of an ideal monatomic gas initially at $1.0 \times 10^{6} P a$ and $300 K$ expands until its volume doubles. Compute the work done if the expansion is isothermal, adiabatic and isobaric.

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Solution

Let $P_{1} = 1.0 \times 10^{6} P a = 10^{6} N / m^{2}$ ; $T_{1} = 300 K$ ; $n = 2 m o l e s$ .
$F i n a l v o l u m e = 2 (i n i t i a l v o l u m e) ⟹ V_{2} = 2 V_{1}$
$W_{(i s o t h e r m a l)} = 2.303 n R T log (\frac{V_{2}}{V_{1}})$
$⟹ W = 2.303 \times 2 \times 8.3 \times 300 log 2$
$W = 3452 J$
$W_{(a d i a b a t i c)} = \frac{n R (T_{1} - T_{2})}{γ - 1}$
For adiabatic process: $T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1}$
$T_{2} = T_{1} = {(\frac{V_{2}}{V_{1}})}^{γ - 1} = 300 \times {(\frac{1}{2})}^{(\frac{5}{3}) - 1} ⟹ T_{2} = 189 K$
$W = \frac{n R (T_{1} - T_{2})}{γ - 1} = \frac{2 \times 8.3 \times (300 - 189)}{(\frac{5}{3}) - 1} ⟹ W = 2764 J$
$W_{(i s o b a r i c)} = P (V_{2} - V_{1}) ⟹ P_{1} V_{1} = n R T_{1}$
$V_{1} = \frac{n R T_{1}}{P_{1}} = \frac{2 \times 8.3 \times 300}{10^{6}} ⟹ V_{1} = 4.98 \times 10^{- 3} m^{3}$
$W = P (V_{2} - V_{1}) = P (2 V_{1} - V_{1}) = P V_{1}$
$⟹ W = 10^{6} \times 4.98 \times 10^{- 3} J ⟹ W = 4980 J$

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