A 120 cm3 box contains helium at a pressure of 2.20 atm and a temperature of 80.

Question

A 120 cm3 box contains helium at a pressure of 2.20 atm and a temperature of 80.0 C. It is placed in thermal contact with a 180 cm3 box containing argon at a pressure of 4.50 atm and a temperature of 420 C.

Part A) What is the initial thermal energy of each gas?

Part B) What is the final thermal energy of each gas?

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

Part D)

Part E) What is the final temperature?

Part F) What is the final pressure in each box?

Explanation / Answer

Thermal energy of an ideal gas is given by:
U = nCvT
(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:
pV = nRT

n = pV/(RT)

-helium
n = pV/(RT)
= 2.2atm *0.12L / ( 0.08205746atmL/molK *(80+273)K)
= 9.11409×10³mol
-argon
n = pV/(RT)
= 4.5atm *0.180L / ( 0.08205746atmL/molK *(420+273)K)
= 14.2441×10³mol

So the initial energies are:
-helium
U = (3/2)nRT
= (3/2) *9.11409×10³mol 8.314472J/molK * (80+273)K
= 40.12489J
-argon
U = (3/2)*n*R*T
= (3/2) *14.2441×10³mol *8.314472J/molK *(420+273)K
= 123.1102J

part 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:
U' + U' = U + U

On the other hand both boxes have same final temperature final T' at equilibrium:

U' = (3/2)nRT'

U' = (3/2)nRT'
hence:
U'/n = U'/n

U' = (n/n) U'

substitute to energy balance

U' + (n/n) U' = U + U

U' = (U + U) / (1 + (n/n))

= (40.12489J + 123.1102J) / (1 + (14.2441×10³mol/9.11409×10³mol) )

= 63.692J

U' = U + U - U'
= 40.12489J + 123.1102J - 63.692J
= 99.54268J

part c &d

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

U = U' - U = 63.692J -40.12 = 23.572J

U > 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.

part e

U' + U' = U + U

(3/2)nRT' + (3/2)nRT' = U + U

T' = (U + U) / {3/2*R(n + n)}

= (40.12489J + 123.1102J) / {(3/2)*8.314472J/molK*(9.11409×10³mol + 14.2441×10³mol)}

= 560.33K
= 287.335°C

part F

since n and V is constant in each box

p/T = nR/V = constant

p' = p(T'/T)

so

p' = p(T'/T)

=2.20atm* (560.33K/ (80+273)K)

= 3.492atm

p' = p(T'/T)
= 4.50atm *(560.33K/ (420+273)K)
= 3.6385atm

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