Explanation:
The given data is as follows.
n = 1 mol,
Q = 1500 J, R = 8.314 J/mol k
(a)
And, according to the first law of thermodynamics
And, in an isothermal process the change in internal energy of the gas is zero.
Hence, 0 = Q - W
or, W = Q
Expression for work done in an isothermal process is as follows.
W =
As W = Q, Hence expression for Q will also be given as follows.
Q =
Now,
[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]
=
= nR ln 2
=
= 5.76 J/K
Therefore, change in entropy is 5.76 J/K.
(b) As, Q =
=
= nRT ln 2
T =
=
= 260.4 K
Therefore, temperature of the gas is 260.4 K.