Calculate the work done when 1 mole of an ideal gas expands from 1dm3
to 10dm3 against a constant
pressure of 1atm. Calculate the work done in dm3
.atm, J and erg
Answers
Answered by
0
Answer:
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 .
Explanation:
Answered by
0
V = nRT / P
= (1 mol × 0.0821 L.atm/(mol K) × 373 K) / (1 atm)
= 30.6 L
Work = P (Vf - Vi)
= 1 atm × (30.6 L - 0 L) ………[∵ When compared to volume of vapour, volume of water is negligible]
= 30.6 L atm
∴ Work done = 30.6 L atm
[Note: In case if you want answer in Joules then convert it by using the relation
1 L atm = 101.3 Joules]
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