Question

A 120 cm3 box contains helium at a pressure of 2.10 atm and a temperature of 100 ∘C. It is placed in thermal contact with a 210 cm3 box containing argon at a pressure of 4.00 atm and a temperature of 390 ∘C.

What is the final thermal energy of each gas?

How much heat energy is transferred, and in which direction?

What is the final temperature?

What is the final pressure in each box?

Answer 1

(a) Thermal energy of an ideal gas is

U = n∙Cv∙T

n number of moles,

C_v molar heat capacity at constant volume ,

T absolute temperature

The molar heat capacity of monoatomic ideal gases like helium and argon is:

Cv = (3/2)∙R

from ideal gas law:

p∙V = n∙R∙T => n = p∙V/(R∙T)

For helium:

n₁ = p₁∙V₁/(R∙T₁)

n₁ = 1.6atm ∙ 0.12L / ( 0.08205746atmL/molK ∙ (100+273)K

n₁=6.27 x 10^-3mol

For Argon

n₂ = p₂∙V₂/(R∙T₂)

n₂ = 4.0atm ∙ 0.210L / ( 0.08205746atmL/molK ∙ (389+273)K)

n₂ =15.44 x 10^-3mol

Initial energy of helium

U₁ = (3/2)∙n₁∙R∙T₁

= (3/2) ∙ 6.27 x 10^-3mol ∙ 8.314472J/molK ∙ (100+273)K

=29.17 J

For argon

U₂ = (3/2)∙n₂∙R∙T₂

= (3/2) ∙ 15.44 x 10^-3mol ∙ 8.314472J/molK ∙ (390+273.15)K

=127.67J

SInce 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   ...............................................(1)

At equilibrium both boxes have same final temperature T'

U₁' = (3/2)∙n₁∙R∙T'

U₂' = (3/2)∙n₂∙R∙T'

so;

U₂' = (n₂/n₁) ∙ U₁'

substitute in equation (1)

U₁' + (n₂/n₁) ∙ U₁' = U₁ + U₂

U₁' = (U₁ + U₂) / (1 + (n₂/n₁))

U₁' = (29.17J + 127.67J) / (1 + (15.44 x 10^-3mol/6.27 x 10^-3mol) )

U₁' = 45.296J ( final thermal energy of helium)

Now,

U₂' = U₁ + U₂ - U₁' = 29.17J + 127.67J - 45.296J = 111.544 J (final thermal energy of Argon)

(b)

The heat energy transferred is equal to the absolute value of thermal energy change in each box:

∆U₁ = U₁' - U₁ = 45.296-29.17 = 16.126 J

∆U₁ > 0 , so the thermal energy of the helium has risen and argon's thermal energy has dropped by the same value. That indicate that heat is transferred from argon to helium.

(c)

U₁' + U₂' = U₁ + U₂

(3/2)∙n₁∙R∙T' + (3/2)∙n₂∙R∙T' = U₁ + U₂

Solving for T'

T' = (U₁ + U₂) / ((3/2)∙R∙(n₁ + n₂))

= (29.17J + 127.67J) / ((3/2)∙8.314472J/molK∙(6.27×10⁻³mol + 15.44×10⁻³mol))

= 579 K (approx)

= 306°C ( final temperature )

(d)

since n and V is constant in each box

p/T = n∙R/V = constant

p' = p∙(T'/T)

p₁' = p₁∙(T'/T₁)

p₁' = 1.6atm ∙ (579K/ (100+273)K) = 2.48 atm

p₂' = p₂∙(T'/T₂)

p₂' = 4.0atm ∙ (579K/ (390+273)K) = 3.4 atm

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