Chemistry, asked by priyanshu5482, 11 months ago

work done during expansion of 5 litre of ideal gas upto 15 litre at constant pressure of 10ATM is​

Answers

Answered by rajkumar707
5

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

Answered by Anonymous
0

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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