Calculate the work done in a reversible expansion of three moles of an ideal gas from pressure of 4 bar to 1 bar at 298 k ?
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Answer:
Hello my friend
Explanation:
n=3 moles, T=27
o
C=300K
P
1
=10atm P
2
=1atm
Work done in isothermal reversible process,
W=−2.303nRTlog(P
1
/P
2
)
=−2.303×3×8.314×300log(
1
10
)
=5744.14J
Work done against constant pressure of 1atm
=P
opp
(V
f
−V
i
)
=−1atm(
P
2
nRT
−
P
1
nRT
)
=−1×3×8.314×300(
1
1
−
10
1
)
=6734.34J .
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