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