First slide

Thermodynamic processes

Question

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $γ$ =1.40, the work done by air is: ( $3^{1.4}$ =4.6555) [ Take air to be an ideal gas)

Moderate

A

100.8J
B

90.5J
C

60.7J
D

48J

Solution

$\begin{array}{l} G i v e n : P_{1} = 1 a t m, T_{1} = 273 K, \frac{V_{1}}{V_{2}} = \frac{1}{3} \\ A d i a b a t i c P r o c e s s, P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \end{array}$
$F i n a l P r e s s u r e, P_{2} = P_{1} {[\frac{V_{1}}{V_{2}}]}^{γ}$ $= 1 a t m {(\frac{1}{3})}^{1.4} = \frac{1}{4.6555} = 0.215 a t m$
Now , work done in adiabatic process
$\begin{array}{l} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} \\ = \frac{(1) (1) - (0.215) (3)}{1.4 - 1} \\ = 0.8875 a t m - l t \\ = 0.8875 \times 1.01325 \times 10^{5} P a \times 10^{- 3} m^{3} \\ = 89.9 J \\ ≃ 90.5 J (c l o s e s t) \end{array}$

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