Question

A 110 cm3 box contains helium at a pressure of 1.60 atm and a temperature of 100 ∘C. It is placed in thermal contact with a 220 cm3 box containing argon at a pressure of 3.60 atm and a temperature of 420 ∘C.

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?

Answer 1

We know that the thermal energy of an ideal gas is given by:

U = n∙Cv∙T

where 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

So,n = p∙V/(R∙T)

For helium

n1 = p1∙V1/(R∙T1)

= 1.6atm ∙ 0.11L / ( 0.08205746atmL/molK ∙ (100+273.15)K)

= 7.86×10^-3mol

For Argon,

n2 = p2∙V2/(R∙T2)

= 3.6atm ∙ 0.22L / ( 0.08205746atmL/molK ∙ (420+273.15)K)

= 13.93×10^-3mol

We can now find the initial energies,So the initial energies are:

For helium

U1 = (3/2)∙n1∙R∙T1

= (3/2) ∙7.86×10^-3mol ∙ 8.314472J/molK ∙ (100+273.15)K

= 26.76 J

Argon

U2 = (3/2)∙n2∙R∙T2

= (3/2) ∙ 15.93×10^-3mol ∙ 8.314472J/molK ∙ (420+273.15)K

= 120.45J

b.We can consider that there 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'

So we have

U1'/n1 = U2'/n2

Rearranging,

U2' = (n2/n1) ∙ U1'

substitute to energy balance

U1' + (n2/n1) ∙ U1' = U1 + U2

Which shows that

U1' = (U1 + U2) / (1 + (n2/n1))

= (26.76J + 120.45J) / (1 + (13.93×10^-3mol/7.86×10^-3mol) )

= 53.097 J

U2' = U1 + U2 - U1'

= 26.76J + 120.45J - 53.097J

= 94.11 J

c.Heat energy transferred = to the absolute value of thermal energy change in each box:

dU1 = U1' - U1 = 53.097J - 26.76 = 26.337 J

dU1 > 0 , so the thermal energy of the helium has risen, while argon's thermal energy has dropped by the same value. which implies that heat energy 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))

= (26.76J + 120.45J) / ((3/2)∙8.314472J/molK∙(7.86×10^-3mol + 15.93×10^-3mol))

= 496.18 K

= 223^oC

e.since n and V is constant in each box

p/T = n∙R/V = constant

p' = p∙(T'/T)

which implies

p1' = p1∙(T'/T1)

= 1.6atm ∙ (496.18K/ (100+273.15)K)

= 2.13 atm

p2' = p2∙(T'/T2)

= 3.6atm ∙ (496.18K/ (420+273.15)K)

= 2.57 atm