A sample of 1 mole of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown. Find change in enthalpy for the cycle as a whole
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In the given question, the process b → c is a isochoric process at a constant volume of 0.800 m3.
a) The work done in a thermodynamic process is given by the equation
W=PΔV=nRΔT.
Since the volume is constant in a isochoric process, ΔV is zero.
Thus the work done by the ideal gas in the process b → c also becomes zero.
b) To calculate the change in internal energy of the gas in the process b → c, we need to find the temperature of the gas at the states b and c.
Let these temperature be Tb and Tc respectively.
According to the ideal gas equation
PV=nRT
Therefore,
T=PVnR.
where
n represents the number of moles of the ideal gas.
R represents the universal gas constant.
T represents the temperature of the gas.
P represents the pressure of the gas.
V represents the volume of the ideal gas.
So the temperature at the states b and c from the graph will be:
T
b
=
P
b
V
b
n
R
=
(
3
×
10
5
P
a
)
(
0.800
m
3
)
n
R
=
240000
n
R
Similarly
T
c
=
P
c
V
c
n
R
=
(
1
×
10
5
P
a
)
(
0.800
m
3
)
n
R
=
80000
n
R
We know that the specific capacity at constant volume of a monoatomic ideal gas is given by
C
V
=
1.5
R
and the change in internal energy Δ U can be given by
Δ
U
=
n
C
V
Δ
T
=
n
(
1.5
R
)
Δ
T
=
1.5
n
R
(
T
f
i
n
a
l
−
T
i
n
i
t
i
a
l
)
.
Thus for the process b → c, the final state is c and the initial state is b.
Hence:
Δ
U
b
c
=
1.5
n
R
(
T
c
−
T
b
)
=
1.5
n
R
(
80000
n
R
−
240000
n
R
)
=
1.5
(
−
160000
)
=
−
240000
J
=
−
240
k
J
The internal energy would decrease by 240 K J by releasing an equal amount of heat to the system.
a) The work done in a thermodynamic process is given by the equation
W=PΔV=nRΔT.
Since the volume is constant in a isochoric process, ΔV is zero.
Thus the work done by the ideal gas in the process b → c also becomes zero.
b) To calculate the change in internal energy of the gas in the process b → c, we need to find the temperature of the gas at the states b and c.
Let these temperature be Tb and Tc respectively.
According to the ideal gas equation
PV=nRT
Therefore,
T=PVnR.
where
n represents the number of moles of the ideal gas.
R represents the universal gas constant.
T represents the temperature of the gas.
P represents the pressure of the gas.
V represents the volume of the ideal gas.
So the temperature at the states b and c from the graph will be:
T
b
=
P
b
V
b
n
R
=
(
3
×
10
5
P
a
)
(
0.800
m
3
)
n
R
=
240000
n
R
Similarly
T
c
=
P
c
V
c
n
R
=
(
1
×
10
5
P
a
)
(
0.800
m
3
)
n
R
=
80000
n
R
We know that the specific capacity at constant volume of a monoatomic ideal gas is given by
C
V
=
1.5
R
and the change in internal energy Δ U can be given by
Δ
U
=
n
C
V
Δ
T
=
n
(
1.5
R
)
Δ
T
=
1.5
n
R
(
T
f
i
n
a
l
−
T
i
n
i
t
i
a
l
)
.
Thus for the process b → c, the final state is c and the initial state is b.
Hence:
Δ
U
b
c
=
1.5
n
R
(
T
c
−
T
b
)
=
1.5
n
R
(
80000
n
R
−
240000
n
R
)
=
1.5
(
−
160000
)
=
−
240000
J
=
−
240
k
J
The internal energy would decrease by 240 K J by releasing an equal amount of heat to the system.
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