Unit 5
The internal energy of an ideal gas
Joule experiment
Setup: Two flasks (A and B) are totally immersed in a water bath, which is
surrounded by a thermally insulated wall. A thermometer is in contact with the water
bath.
flask A contains gas at high pressure and separated by valve (in closed
position) from flask B, which is completely evacuated
Valve is opened, and gas would expand against the vacuum while
temperature is monitored
Result: No temperature change was observed
Conclusion: q= 0 and w = 0 (dw = -PextdV , Pext = 0)
ΔU = q + w = 0
(1)
A total differential (dU) can be setup for this system that evaluated volume change
and temperature change:
𝜕𝑈 𝜕𝑈
𝑑𝑈 = ( ) 𝑑𝑉 + ( ) 𝑑𝑇 = 0
𝜕𝑉 𝑇 𝜕𝑇 𝑉
(2)
, Effect on U when
Step 1: changing V while keeping T constant
Step 2: changing T while keeping V constant
U
(1) dU = 0
(2) T
V
Rearrangement of equation 2:
𝜕𝑈 𝜕𝑈 𝜕𝑇
( ) = −( ) ( )
𝜕𝑉 𝑇 𝜕𝑇 𝑉 𝜕𝑉 𝑈
(3)
From Joule experiment: temperature did not change on expansion of gas, therefore
𝜕𝑇
( ) =0
𝜕𝑉 𝑈
This implies from equation 3 that:
𝜕𝑈
( ) =0
𝜕𝑉 𝑇
For an ideal gas, the internal energy is independent of volume.
𝜕𝑈
( ) = 𝜋𝑇
𝜕𝑉 𝑇
𝜋 𝑇 = internal pressure= 0 for ideal gas
The internal energy of an ideal gas
Joule experiment
Setup: Two flasks (A and B) are totally immersed in a water bath, which is
surrounded by a thermally insulated wall. A thermometer is in contact with the water
bath.
flask A contains gas at high pressure and separated by valve (in closed
position) from flask B, which is completely evacuated
Valve is opened, and gas would expand against the vacuum while
temperature is monitored
Result: No temperature change was observed
Conclusion: q= 0 and w = 0 (dw = -PextdV , Pext = 0)
ΔU = q + w = 0
(1)
A total differential (dU) can be setup for this system that evaluated volume change
and temperature change:
𝜕𝑈 𝜕𝑈
𝑑𝑈 = ( ) 𝑑𝑉 + ( ) 𝑑𝑇 = 0
𝜕𝑉 𝑇 𝜕𝑇 𝑉
(2)
, Effect on U when
Step 1: changing V while keeping T constant
Step 2: changing T while keeping V constant
U
(1) dU = 0
(2) T
V
Rearrangement of equation 2:
𝜕𝑈 𝜕𝑈 𝜕𝑇
( ) = −( ) ( )
𝜕𝑉 𝑇 𝜕𝑇 𝑉 𝜕𝑉 𝑈
(3)
From Joule experiment: temperature did not change on expansion of gas, therefore
𝜕𝑇
( ) =0
𝜕𝑉 𝑈
This implies from equation 3 that:
𝜕𝑈
( ) =0
𝜕𝑉 𝑇
For an ideal gas, the internal energy is independent of volume.
𝜕𝑈
( ) = 𝜋𝑇
𝜕𝑉 𝑇
𝜋 𝑇 = internal pressure= 0 for ideal gas
