Answer : The change in entropy is 6 J/K
Explanation :
To calculate the change in entropy we use the formula:
[tex]\Delta S=\int \frac{dQ}{T}[/tex]
and,
[tex]Q=nC_pdT[/tex]
[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p}dT}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
n = number of moles = 2 moles
[tex]T_f[/tex] = final temperature = 303 K
[tex]T_i[/tex] = initial temperature = 273 K
[tex]C_{p}[/tex] = heat capacity at constant pressure = [tex]0.1\times T(J/K.mol)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=2\int\limits^{303}_{273}{\frac{(0.1\times TdT}{T}[/tex]
[tex]\Delta S=2\times 0.1\int\limits^{303}_{273}dT[/tex]
[tex]\Delta S=2\times 0.1\times [T]^{303}_{273}[/tex]
[tex]\Delta S=2\times 0.1\times (T_f-T_i)[/tex]
[tex]\Delta S=2\times 0.1\times (303-273)[/tex]
[tex]\Delta S=6J/K[/tex]
Therefore, the change in entropy is 6 J/K
