Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 0 25 C into a vacuum until its total volumes is 10 litres. How much heat is absorbed and how much work is done the expansion?
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(i) In case of expansion, work is done by the system.
Now, as we know that,
W=−P
ext.
ΔV
As the gas is expanding is vaccum which has no pressure, i.e.,
p
ext.
=0
∴W=0
Hence no work is done.
From first law of thermodynamics,
ΔU=q+W
As the system is working at constant temperature, i.e., isothermally.
∴ΔU=0
Hence q=−W=0
Hence no heat is absorbed in the expansion.
(ii) W=−P
ext.
ΔV
P
ext
=1atm
ΔV=10−2L=8L
∴W=−1×8=−8atm−L
As, q=−W=−(−8)J=8atm−L
(iii) Now, as the work is done reversibly, then the work done will be given as-
W
rev.
=−2.303nRTlog
V
i
V
f
⇒W
rev.
=−2.303PVlog
2
10
⇒W
rev.
=−2.303×1×10×log5=−16.1L−atm[∵Given P=1 atm in (ii) part]
As q
rev.
=−W
rev.
q
rev.
=16.1L−atm
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