Question

Review A 110 cm3 box contains helium at a pressure of 2.50 atm and a temperature of 120 ∘C. It is placed in thermal contact with a 180 cm3 box containing argon at a pressure of 3.60 atm and a temperature of 420 ∘C.  

What is the initial thermal energy of each gas? Correct answer: 41.8 J for He and 98.5 J for Ar

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

1)

First i would compute the amount of gas in each box. From ideal gas law follows

n = P∙V/(R∙T)

In the helium box are:

n₁ = P₁∙V₁/(R∙T₁)

= 2.5∙101325 Pa ∙ 110×10⁻⁶ m³ / ( 8.3145 Jmol⁻¹K⁻¹ ∙ (120 + 273) K ) = 8.53×10⁻³ mol

The amount of argon is:

n₂ = P₂∙V₂/(R∙T₂)

= 3.6∙101325 Pa ∙ 180×10⁻⁶ m³ / ( 8.3145 Jmol⁻¹K⁻¹ ∙ (420 + 273) K ) = 11.4×10⁻³ mol

The internal energy of an ideal gas is given by:

U = n∙ Cv ∙ T

The molar heat capacity at constant volume for monatomic ideal gases, like Helium and Argon is:

Cv = (3/2)∙R

So the initial internal energies in the boxes are:

U₁ = 8.53×10⁻³ mol ∙ (3/2) ∙ 8.3145 Jmol⁻¹K⁻¹ ∙ (120 + 273) K = 41.80 J

U₂ = 11.4×10⁻³ mol ∙ (3/2) ∙ 8.3145 Jmol⁻¹K⁻¹ ∙ (420 + 273) K = 98.5 J

(2) The total internal energy is:

Ut = U₁ + U₂ = 140.3 J

Assuming there is no heat loss to the surrounding, the total internal energy in the boxes stays constant. Thus after the equilibrium has established:

Ut = U₁' + U₂'

At equilibrium gas has the same temperature in both boxes. So both T and Cv are same in final state and the thermal energy in each box is determined by the number moles in it. Hence the ratio of the internal energies is equal to ration of number of moles:

U₁'/U₂' = n₁/n₂

or

U₂' = U₁' ∙(n₂/n₁)

Hence,

Ut = U₁' + U₁' ∙(n₂/n₁)

=>

U₁' = Ut/(1 + (n₂/n₁))

= 140.3 J / (1 + (11.4×10⁻³ mol /8.53×10⁻³ mol))

= 60.0 J

=>

U₂' = Ut - U₁' = 140.3 J - 60.0 J = 80.3 J

3)

The magnitude of change in internal is the same for both boxes (just the sign is different). The energy of the helium box rises as it absorbs heat from the argon box

Q = ΔU₁ = U₁' - U₁ = 60 J - 41.80 J = 18.2 J

4)

From relation above follows for the change in internal energy in the helium box:

ΔU₁ = n₁∙(3/2)∙R∙ΔT₁

So the change in temperature is:

ΔT₁ = ΔU₁ / (n₁∙(3/2)∙R )

= 18.2 J / (8.53×10⁻³ mol ∙ (3/2) ∙ 8.3145 Jmol⁻¹K⁻¹ )

= 171.0 K = 171.0 °C

So the final of the helium box and the argon box is;

T₂' = T₁' = T₁ + ΔT₁ = 120°C + 171 °C = 291 °C

5)

Note that both volume and amount fo gas in each box remains unchanged. Therefore

P/T = n∙R/V = constant

or

P'/T' = P/T

<=>

P' = P ∙ (T'/T)

=>

P₁' = P₁ ∙ (T₁'/T₁) = 2.5 atm ∙ ( (291 + 273) / (120 + 273) ) = 3.59 atm

P₂' = P₂ ∙ (T₂'/T₂) = 3.6 atm ∙ ( (291 + 273) / (420 + 273) ) = 2.93 atm