Question

A fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from $-{73}^{\circ }\text{C}$ to ${27}^{\circ }\text{C}$. Calculate the change in Gibbs energy if the entropy of the gas is a function of temperature given as, $S=2+{10}^{-2}\text{T (J/K)}:(\text{Use 1 atm - L = 0.1 kJ}$)

• A. $1231.5\text{J}$
• B. $1281.5\text{J}$
• C. $781.50\text{J}$
• D. $0\text{J}$

Answer 1

The correct option is C $781.50\text{J}$

Given data:

$V=24.63L$

$T_1=-73+273=200K$

$P_1=1atm$

$T_2=27+273=300K$

$S=2+{10}^{-2}J/K$

We know that:

$dG=VdP-SdT$

First we calculate $p_2$

At constant volume

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{3}{2}atm$

$\int dG=V{\int }_1^{3/2}dP-{\int }_{200}^{300}(2+{10}^{-2}T)dT$

$=24.63\times 0.5\times 100-|2T+{10}^{-2}\times \frac{T^2}{2}{|}_{200}^{300}$

$=781.5J$