Question

2.5 g of helium at an initial temperature of 310 K interacts thermally with 7.0 g of oxygen at an initial temperature of 640 K

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

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

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

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

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

D) What is the final temperature?

Answer 1

Given,

mHe = 2.5 g ; mO = 7 g ; KiHe = 310 K ; KiO = 640 K

A)The intial energy of a monoatomic gas is:

Ui(He) = 3/2NkT

Ui(He) = 1.5 x 2/4 x 6.023 x 10^23 x 1.38 x 10^-23 x 310 = 1932.48 J

UiO = 5/2NkT

Ui(O) = 2.5 x 8/32 x 6.02 x 10^23 x 1.38 x 10^-23 x 640 = 3323.04 J

Hence, Ui(He) = 1932.48 J ; Ui(O) = 3323.04 J

B)From conservation of enegy

Intial energy = Final energy

Ui(He) = 1932.48 J ; Ui(O) = 3323.04 J

After the equilibrium reaches both the gases will be at common temp.

3323.04 + 1932.48 = 5255.52 = 3/2 N k T + 5/2 N k T

T (3 x 2/4 x 6.023 x 10^23 x 1.38 x 10^-23 + 5 x 8/32 x 6.02 x 10^23 x 1.38 x 10^-23) = 10511.04

T = 10511.04/22.85 = 460 K

Uf(He) = 1.5 x 2/4 x 6.023 x 10^23 x 1.38 x 10^-23 x 460 = 2867.55 J

Ui(O) = 2.5 x 8/32 x 6.02 x 10^23 x 1.38 x 10^-23 x 460 = 2388.44 J

Hence, Uf(He) = 2867.55 J ; Uf(O) = 2388.44 J

C)Heat is transferred from the gas at higher temperature to the one at lower temp.

So, from O --> He

Q(trans) = 3323.04 - 2388.44 = 934.6 J

Hence, Q(trans) = 934.6 J

D)T = 460 K (calculated in B)