Answer:
a) W = 900
J. b) Q = 3142.8
J
. c) ΔU = 2242.8
J. d) W = 0. e) Q =
2244.78 J. g) Δ
U = 0.
Explanation:
(a) Work done by the gas during the initial expansion:
The work done W for a thermodynamic constant pressure process is given as;
W = p
Δ
V
where
p is the pressure and Δ
V is the change in volume.
Here, Given;
P
1
=
i
n
i
t
i
a
l p
r
e
s
s
u
r
e = 2.5
×
10^
5
P
a
T
1
=
i
n
i
t
i
a
l
t
e
m
p
e
r
a
t
u
r
e = 360
K
n
=
n
u
m
b
er
o
f
m
o
l
e
s = 0.300 m
o
l
The ideal gas equation is given by
P
V = nRT
where
,
p = absolute pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = universal gas constant = 8.314
K
J
/
m
o
l
K
T = absolute temperature of the gas
Now we will Calculate the initial volume of the gas using the above equation as follows;
PV = n
R
T
2.5
×
10
^5
×
V
1 = 0.3
×
8.314
×
360
V1 = 897.91 / 250000
V
1 = 0.0036
m
^3 = 3.6×10^-3 m^3
We are also given that
V
2 = 2×
V
1
V2 = 2
×
0.0036
V2 = 0.0072
m^3
Thus, work done is calculated as;
W = p
Δ
V = p×(V2 - V1)
W = (
2.5
×
10
^5
)
×(
0.0072 − 0.0036
)
W = 900
J.
(b) Heat added to the gas during the initial expansion:
For a diatomic gas,
C
p = 7
/2
×R
Cp = 7
/2
×
8.314
Cp = 29.1 J
/
mo
l
K
For a constant pressure process,
T
2
/T
1 = V
2
/V
1
T
2 = V
2
/V
1
×
T
1
T
2 = 2
×
T
1 = 2×360
T
2 = 720 K
Heat added (Q) can be calculated as;
Q = n
C
p
Δ
T = nC×(T2 - T1)
Q = 0.3
×
29.1
×
(
720 − 360
)
Q = 3142.8
J
.
(c) Internal-energy change of the gas during the initial expansion:
From first law of thermodynamics ;
Q = Δ
U
+
W
where
,
Q is the heat added or extracted,
Δ
U is the change in internal energy,
W is the work done on or by the system.
Put the previously calculated values of Q and W in the above formula to calculate Δ
U as;
Δ
U = Q − W
ΔU = 3142.8 − 900
ΔU = 2242.8
J.
(d) The work done during the final cooling:
The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.
(e) Heat added during the final cooling:
The final process is a isochoric process and for this, the first law equation becomes
,
Q = Δ
U
The molar specific heat at constant volume is given as;
C
v = 5
/2
×R
Cv = 5
/2
×
8.314
Cv = 20.785 J
/
m
o
l K
The change in internal energy and thus the heat added can be calculated as;
Q =
Δ
U = n
C
v
Δ
T
Q = 0.3
×
20.785
×
(
720 - 360
)
Q =
2244.78 J.
(f) Internal-energy change during the final cooling:
Internal-energy change during the final cooling is equal to the heat added during the final cooling Q = Δ
U .
(g) The internal-energy change during the isothermal compression:
For isothermal compression,
Δ
U = n
C
v
Δ
T
As their is no change in temperature for isothermal compression,
Δ
T
=
0
, then,
Δ
U = 0.
. The breakdown voltage of dry air is about 3x10^6V/m. So solving for V we get
