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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?

Respuesta :

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

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