work done during expansion of 5 litre of ideal gas upto 15 litre at constant pressure of 10ATM is
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Answer:
Work done under constant pressure
W = PΔV
Here 1atm = 10^5 Pa, 1 litre = 10^-3 m^3
W = (10*10^5)*(15 - 5)*10^-3 = 10^4J
W = 10000J = 10^4J
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Answer:
here is ur answer......
Explanation:
In irreversible, work done is W= -P(V2 -V1) So work done is 5*3=15 litre atm or (15*101) Joule.......
....i hope it had helped u....
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