2 mole of an ideal gas at temperature 28 degrees celcios and 2 atm pressure is compressed to half of its volume by applying a constant pressure up to 4 atm. then calculate work done on or by the system and heat absorbed or released by the gas.
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Answer:
To calculate the work done, We need to know the final temperature.
T
1
=300K,P
1
=1 atm,n=2 mole,P
2
=2 atm,V
1
=
P
1
nRT
1
w=ΔU=nC
v
(T
2
−T
1
)=−P
2
(V
2
−V
1
)
For a monatomic gas, C
v
=
2
3
R
⇒
2
3
nR(T
2
−T
1
)=−P
2
(V
2
−V
1
)
2
3
nR(T
2
−T
1
)=−P
2
(nR
P
2
T
2
−nR
P
1
T
1
)
2
3
(T
2
−T
1
)=−T
2
+
P
1
P
2
⇒T
2
=
5
2
(
P
1
P
2
+
2
3
)T
1
T
2
=0.4(2+1.5)300=420K
Work done, W=nC
v
ΔT=2×
2
3
R×(420−300)
⇒W=2×
2
3
×2×120=720 cal
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