10 g-mol an ideal gas follows the cyclic path as shown in the figure. At initial state A pressure is 10 atm and volume is 10 liter. Pressure and volume at point C are 1 atm and 50 liter respectively. Calculate (i) net work done in atm-liter and (ii) temperature at states A, B, C and D in oC
Answers
Answer:
We know,
Path CA-Isothermal compression
Path AB-Isobaric expansion
Path BC=Isochoric change
Let V
i
and V
f
are initial volume and final volume at respective points.
For temperature T
1
(For C):PV=nRT
1
2×10=1×0.0821×T
1
∴T
1
=243.60K
For temperature T
2
(ForCandB):
T
1
P
1
V
1
=
T
2
P
2
V
2
T
1
2×10
=
T
2
20×10
∴
T
1
T
2
=10
∴T
2
=243.60×10=2436.0K
PathCA: ΔU=0 for isothermal compression; Also q=w
w=+2.303nRT
1
log
V
f
V
i
=2.303×1×2×243.6×log
1
10
=+1122.02cal
PathAB:w=−P(V
f
−V
i
)
w=−20×(10−1)=−180litre−atm
=
0.0821
−180×2
=−4384.9cal
PathBC:w=−P(V
f
−V
i
)=0
(∴V
f
−V
i
=0;sincevolumeisconstant)
For monoatomic gas heat change at constant volume=q
r
=ΔU
Thus for path BC q
r
=C
v
×n×ΔT=ΔU
q
r
=
2
3
R×1×(2436−243.6)
2
3
×2×1×2192.4
=6577.2cal
Since, process involve cooling∴q
v
=ΔU=−6577.2cal
Also in path AB, the internal energy in state A and state C is same. Thus during path AB, an increase in internal energy equivalent of change in internal energy during path BC should take place.Thus,ΔU for path AB=+6577.2cal
Now q for path AB=ΔU−w
AB
=6577.2+4384.9=10962.1cal
Cycle:ΔU=0:q=−w=−[w
PathCA
+w
PathAB
+w
PathBC
]
=−[+1122.02+(−4384.9)+0]
∴q=−w=+3262.88cal
Answer:
10 g-mol an ideal gas follows the cyclic path as shown in the figure. At initial state A pressure is 10 atm and volume is 10 liter. Pressure and volume at point C are 1 atm and 50 liter respectively. Calculate (i) net work done in atm-liter and (ii) temperature at states A, B, C and D in oC