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.]

The final temperature at the equilibrium is 337.1 K

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