If a gas at 5atm and 373K expands against a constant external pressure of 1atm from a volume of 2 litres to 10 litres, the amount of work done is:
(Given: 1L.atm=101.3J)
A
−402J
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B
−525.3J
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C
−212.8J
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D
−810.4J
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Solution
The correct option is D−810.4J As the process is done at a constant external pressure, it is irreversible. W=−PΔV ⇒W=−1atm[10L−2L] ⇒W=−8L.atm
As, 1L.atm=101.3J
i.e W=−8×101.3J=−810.4J