Question

Consider a rigid, insulated box containing 0.400 mole of He(g) at 24.4°C and 1.00 atm in one compartment and 0.600 mole of N2(g) at 80.1°C and 2.00 atm in the other compartment. These compartments are connected by a partition that transmits heat. What is the final temperature in the box at thermal equilibrium? [For He(g), Cv = 12.5 J K-1 mol-1; for N2(g), Cv = 20.7 J K-1 mol-1.]

Answer 1

The final temperature at the equilibrium is 337.1 K

Step-by-step explanation:

Step 1: Data given

Number of moles He = 0.400 moles

Temperature = 24.4 °C = 297.4 K

Number of moles N2 = 0.600 moles

Temperature = 80.1 °C = 353.1 K

Pressure is 2.00 atm

For He(g), Cv = 12.5 J/K*mol

for N2(g), Cv = 20.7 J/K*mol

Step 2: Calculate the final temperature

Q(He) = -Q(N2)

Q = n * Cv * ΔT

n(He)*Cv(He) * ΔT(He) = - n(N2)*Cv(N2) * ΔT(N2)

⇒with n(He) = the number of moles of Helium = 0.400 moles

⇒with Cv(He) = 12.5 J/K*mol

⇒with ΔT(He) = the change in temperature = T2 - T1 = T2 - 297.4 K

⇒with n(N2) = the number of moles N2 = 0.600 moles

⇒with Cv(N2) =20.7 J/K*mol

⇒with ΔT(N2) = the change in temperature = T2 - T1 = T2 - 353.1 K

0.400 * 12.5 * (T2 - 297.4) = - 0.600 * 20.7 * (T2 - 353.1)

5 * (T2 - 297.4) = -12.42 * (T2 - 353.1)

5T2 - 1487 = -12.42T2 + 4385.5

17.42T2 = 5872.5

T2 = 337.1 K

The final temperature at the equilibrium is 337.1 K