40J heat is supplied to a gaseous system. Simultaneously the gas expands against external pressure of 1atm by 0.5L calculate change in internal energy of the system
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a
-40J
b
40J
c
+10.63J
d
-10.63J
answer is D.
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Detailed Solution
According to first law of thermodynamicsΔU=Q+WQ=+40J;W=-PextΔV=-1×0.5=0.5 Latm W=-50.63J(∴1 Latm =101.26J)ΔU=+40-(50.63) =-10.63J