First slide
Work done by gas
Question

Air expands from 5 litres to 10 litres at 2 atm pressure. External workdone is

Easy
Solution

{V_1} = 5lts = 5 \times {10^{ - 3}}{m^3};{V_2} = 10lts = 10 \times {10^{ - 3}}{m^3}

P = 2atm = 2 \times 1.01 \times {10^5}pa \approx 2 \times {10^5}p.a

\Delta W = P\Delta V = p\left( {{v_2} - {v_1}} \right)

= 2 \times {10^5}(10 \times {10^{ - 3}} - 5 \times {10^{ - 3}})J

= 2 \times {10^5} \times 5 \times {10^{ - 3}} = 1000J

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