Question

A 110cm3 box contains helium at a pressure of 2.00atm and a temperature of 100?C. It is placed in thermal contact with a 210cm3 box containing argon at a pressure of 4.00atm and a temperature of 420?C.

What is the initial thermal energy of each gas?

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
U = n*Cv*delta T
Cv = (3/2)*R            (for helium or argon)
n = p*V/(R*T)

for helium
n1 = p1*V1/(R*T1)
= 2 atm * 0.11L / ( 0.08205atmL/molK * (373.15)K)
= 7.18*10^-3 mol
for argon
n2 = p2*v2/(R*T2)
= 4.0atm * 0.21L / ( 0.08205atmL/molK * (693.15)K)
= 14.76*10^-3 mol

So initial energies are:
for helium
U1= (3/2)*n1*R*T1
= (3/2) *7.18*10^-3 mol * 8.314J/molK * (373.15)K
= 33.41 J
for argon
U2= (3/2)*n2*R*T2
= (3/2) * 14.76*10^-3 mol * 8.314J/molK *(693.15)K
= 127.58 J
b. the sum of the final thermal energies equals the sum of the initial energies:
U1' + U2' = U1 + U2

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))
= (33.41 J + 127.58 J) / (1 + (14.76*10^-3mol/(7.18*10^-3mol)) )
= 52.68 J

U2' = U1 + U2 - U1'
= 33.41J + 127.58J - 52.68J
= 108.31 J
c. The heat energy transferred is
delta U1 = U1' - U1 = 52.68J - 33.41 = 19.27 J

delta 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' = U? + U?
T' = (U? + U?) / ((3/2)*R*(n1 + n2))
= (33.41J + 127.58J) / ((3/2)*8.314J/molK*(7.18*10^-3mol + 14.76*10^-3mol))
= 588.38K
= 315.23 deg C
e.   n and V is constant in each box
p/T = n*R/V = constant
p' = p*(T'/T)
p1' = p1*(T'/T1)
= 2atm * (588.38K/ (373.15)K)
= 3.15 atm = 319173.75 Pa
p2' = p2*(T'/T2)
= 4.0atm * (588.38K/ (693.15)K)
= 3.4 atm = 344505 Pa