56 grams of nitrogen gas expands at 137o C reversibly from 9 liters to 27 liters. Determine q, w, ∆U and ∆H, assuming nitrogen gas behaves ideally at 137o C.
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Since the external pressure is greatly different from the pressure of
N
2
and thus, the process is irreversible.
w=−P
ext
(V
2
−V
1
)
w=−1×(V
2
−V
1
)
Given V
1
=2 litre V
2
=? T=273K
P
1
=5atm P
2
=1atm
∴P
1
V
1
=P
2
V
2
∴V
2
=
1
2×5
=10 litre
∴w=−1×(10−2)=−8 litre atm
∴=−
0.0821
8×1.987
calorie=−
0.0821
8×1.987×4.184
J=−810.10 joule
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