Question

Physics Question!!!!!!!!​

Answer 1

Q] ______?

Solution:

In a adiabatic process,

Q = 0

So, From the first law of thermodynamics,

∆Q = 0

W = -∆U = -nCv∆T

W = -n [ R / ɣ - 1 ] ( Ti - Tf )

Given, work done, W = 6R J, n = 1 mol,

R = 8.31 J / mol-K, ɣ = 5/3, Ti = TK

Substituting given values in Eq [i] we get,

∴ 6 R = [ R / ( 5/3 - 1 ) ] ( T - Tf )

=> 6R = 3R / 2 ( T - Tf )

=> T - Tf = 4

∴ Tf = ( T - 4 ) K

The Final temperature of the gas will be (Tf ) = ( T - 4 ) K !

Answer 2

Answer:

(T+882j

Explanation:

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