Respuesta :
Explanation:
The given data is as follows.
n = 1 mol, [tex]V_{f} = 2V_{i}[/tex]
Q = 1500 J, R = 8.314 J/mol k
(a) [tex]\Delta S = \frac{dQ}{dT}[/tex]
And, according to the first law of thermodynamics
[tex]\Delta E_{int} = Q - W[/tex]
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 = [tex]nRT ln \frac{V_{f}}{V_{i}}[/tex]
As W = Q, Hence expression for Q will also be given as follows.
Q = [tex]nRT ln \frac{V_{f}}{V_{i}}[/tex]
Now,
[tex]\Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}[/tex]
[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]
= [tex]nR ln \frac{2V_{i}}{V_{i}}[/tex]
= nR ln 2
= [tex]1 \times 8.314 \times 0.693[/tex]
= 5.76 J/K
Therefore, change in entropy is 5.76 J/K.
(b) As, Q = [tex]nRT ln \frac{V_{f}}{V_{i}}[/tex]
= [tex]nRT ln \frac{2V_{i}}{V_{i}}[/tex]
= nRT ln 2
T = [tex]\frac{Q}{nR ln 2}[/tex]
= [tex]\frac{1500}{1 \times 8.314 ln 2}[/tex]
= 260.4 K
Therefore, temperature of the gas is 260.4 K.
