Answer:
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
