A 120cm^3 box contains helium at a pressure of 1.60atm and a temperature of 120C. It is placed in thermal contact with a 200cm^3 box containing argon at a pressure of 4.0atm and a temperature of 380C.
a. What is the initial thermal energy of each gas?
b. What is the final thermal energy of each gas?
c. How much heat energy is transferred, and in which direction?
d. What is the final temperature?
e. What is the final pressure in each box?
2 Answers

a.
Thermal energy of an ideal gas is given by:
U = n∙Cv∙T
(n number of moles, Cv molar heat capacity at constant volume , T absolute temperature)
The molar heat capacity of monatomic ideal gases like helium and argon is:
Cv = (3/2)∙R
The number of moles of each gas can be found from ideal gas law:
p∙V = n∙R∙T => n = p∙V/(R∙T)
helium
n₁ = p₁∙V₁/(R∙T₁)
= 1.6atm ∙ 0.12L / ( 0.08205746atmL/molK ∙ (120+273.15)K)
= 5.95×10⁻³mol
argon
n₂ = p₂∙V₂/(R∙T₂)
= 4.0atm ∙ 0.2L / ( 0.08205746atmL/molK ∙ (380+273.15)K)
= 14.93×10⁻³mol
So the initial energies are:
helium
U₁ = (3/2)∙n₁∙R∙T₁
= (3/2) ∙ 5.95×10⁻³mol ∙ 8.314472J/molK ∙ (120+273.15)K
= 29.18J
argon
U₂ = (3/2)∙n₂∙R∙T₂
= (3/2) ∙ 14.93×10⁻³mol ∙ 8.314472J/molK ∙ (380+273.15)K
= 121.59J
b.
Assuming threre is only energy exchanged between the boxes, the total internal energy is conserved. So the sum of the final thermal energies equals the sum of the initial energies:
U₁’ + U₂’ = U₁ + U₂
On the other hand both boxes have same final temperature final T’ at equilibrium:
U₁’ = (3/2)∙n₁∙R∙T’
U₂’ = (3/2)∙n₂∙R∙T’
hence:
U₁’/n₁ = U₂’/n₂
<=>
U₂’ = (n₂/n₁) ∙ U₁’
substitute to energy balance
U₁’ + (n₂/n₁) ∙ U₁’ = U₁ + U₂
=>
U₁’ = (U₁ + U₂) / (1 + (n₂/n₁))
= (29.18J + 121.59J) / (1 + (14.93×10⁻³mol/5.95×10⁻³mol) )
= 42.98J
=>
U₂’ = U₁ + U₂ – U₁’
= 29.18J + 121.59J – 42.98J
= 107.79J
c.
The heat energy transferred is equal to the absolute value of thermal energy change in each box:
∆U₁ = U₁’ – U₁ = 42.98J – 29.18 = 13.8J
∆U₁ > 0 , so the thermal energy of the helium has risen, while argon’s thermal energy has dropped by the same value. That means heat is transferred from argon to helium.
d.
U₁’ + U₂’ = U₁ + U₂
<=>
(3/2)∙n₁∙R∙T’ + (3/2)∙n₂∙R∙T’ = U₁ + U₂
=>
T’ = (U₁ + U₂) / ((3/2)∙R∙(n₁ + n₂))
= (29.18J + 121.59J) / ((3/2)∙8.314472J/molK∙(5.95×10⁻³mol + 14.93×10⁻³mol))
= 579K
= 306°C
e.
since n and V is constant in each box
p/T = n∙R/V = constant
<=>
p’ = p∙(T’/T)
=>
p₁’ = p₁∙(T’/T₁)
= 1.6atm ∙ (579K/ (120+273.15)K)
= 2.356atm
p₂’ = p₂∙(T’/T₂)
= 4.0atm ∙ (579K/ (380+273.15)K)
= 3.546atm

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