Respuesta :
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
We know that,
The relation between the [tex]C_p\text{ and }C_v[/tex] for an ideal gas are :
[tex]C_p-C_v=R[/tex]
As we are given :
[tex]C_p=28.253J/K.mole[/tex]
[tex]28.253J/K.mole-C_v=8.314J/K.mole[/tex]
[tex]C_v=19.939J/K.mole[/tex]
Now we have to calculate the entropy change of the gas.
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
[tex]\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J[/tex]
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. [tex](w=-pdV)[/tex]
(C) Heat during the process will be,
[tex]q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J[/tex]
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
