A container is closed by a movable piston w
supply heat to gas inside container. Duet
heating, piston is displaced by 5 cm. Croes
section area of cylinder is 40 cm and heat supplied
to gas is 40 joules. The change in internal
energy is
Answers
Answer:
Explanation:
.
The temperature must go up if the piston remains stationary
If the piston remains stationary, we know that
Δ
V
, the change in volume of the system, is necessarily zero, i.e.
Δ
V
=
0
.
This implies that the system does no work on the environment as it does not expand.
W
sys
=
∫
P dV
where
P
is the pressure of the system and
V
is the volume
By the first law of thermodynamics:
Q
in
=
W
out
+
Δ
E
th
If no work is done by the system,
W
out
=
0
, so the first law statement becomes:
⇒
Q
in
=
Δ
E
th
If there is no change in the temperature of the gas,
Q
and
Δ
E
would be zero as well, which would not be possible for the situation.
Δ
E
=
3
2
N
k
b
Δ
T
We can therefore convince ourselves that we must have a temperature change.
Assuming an ideal gas, we should be able to apply the ideal gas law:
P
V
T
=
constant
i.e.
P
1
V
1
T
1
=
P
2
V
2
T
2
Since
V
1
=
V
2
, this simplifies to:
P
1
T
1
=
P
2
T
2
For the ideal gas law to be valid, we must have that if the pressure decreases, the temperature decreases by the same amount, and vice versa.
It doesn't make much sense intuitively for the temperature of the gas to decrease, which would mean that the pressure also decreases, as long as we have the constant heat source and no work being done. This is also clear from our statement of
Q
in
, which is a positive quantity as heat is being added to the system. Therefore
Δ
E
must also be a positive quantity. Since
Δ
E
=
3
2
N
k
b
Δ
T
⇒
=
3
2
N
k
b
(
T
f
−
T
i
)
and
N
and
k
b
remain constant,
T
f
>
T
i
.
So we can conclude that the temperature must increase.
Answer:
hope it HELPS
MARK AS BRAINLIEST
