A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $\gamma$ = 1.40, the work done by air is : (31.4 = 4.6555) [Take air to be an ideal gas]
Solution
${P_1}V_1^\gamma = {P_2}V_2^\gamma$
<br><br>$\Rightarrow$ P<sub>2</sub> = P<sub>1</sub>${\left[ {{{{V_1}} \over {{V_2}}}} \right]^\gamma }$
<br><br>= 10<sup>5</sup> $\times$ ${\left[ {{1 \over 3}} \right]^{1.4}}$
<br><br>Work done = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$
<br><br>= $${{{{10}^5} \times {{10}^{ - 3}} - {{{{10}^5}} \over {{3^{1.4}}}} \times 3 \times {{10}^{ - 3}}} \over {1.4 - 1}}$$
<br><br>= 88.7 J $\approx$ 90.5 J
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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