4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the process of coming to thermal equilibrium it transfers 600 J of heat energy to gas B.

What is the initial thermal energy of B?

14657.32 J

Explanation:

Given Parameters;

Number of moles mono atomic gas A, n 1 = 4.2 mol

Number of moles mono atomic gas B, n 2 = 3.2mol

Initial energy of gas A, K A = 9500 J

Thermal energy given by gas A to gas B, Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2) n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3) / (2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2) n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [ (3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J