Fixed amount of an ideal gas contained in a sealed rigid vessel (V = 24.6 litre) at 1.0 bar is heated reversibly from 27°C to 127°C. Determine change in Gibb's energy (in Joule) if entropy of gas S = 10 + 10–2 T (J/K).

To solve this problem, we need to use the Gibbs-Helmholtz equation, which relates the change in Gibbs free energy (ΔG) to the change in enthalpy (ΔH), the change in entropy (ΔS), and the final temperature (Tf).

The equation is: ΔG = ΔH - TfΔS

First, we need to calculate the change in entropy (ΔS). The entropy of the gas is given by the equation S = 10 + 10^-2T. We can find ΔS by subtracting the initial entropy (Si) from the final entropy (Sf).

The initial temperature (Ti) is 27°C, which is 300.15K in Kelvin. The final temperature (Tf) is 127°C, which is 400.15K in Kelvin.

So, Si = 10 + 10^-2 * 300.15 = 13 J/K And, Sf = 10 + 10^-2 * 400.15 = 14 J/K

Therefore, ΔS = Sf - Si = 14 J/K - 13 J/K = 1 J/K

Next, we need to calculate the change in enthalpy (ΔH). For an ideal gas in a sealed rigid vessel, the change in enthalpy is equal to the heat added to the system, which is given by the equation Q = nCΔT, where n is the number of moles, C is the heat capacity, and ΔT is the change in temperature.

We don't have the values of n and C, but we know that the volume (V) is constant and the pressure (P) is 1.0 bar, which is 100000 Pa in Pascal. So, we can use the ideal gas law, PV = nRT, to find that n = PV/RT = 100000 * 24.6 / (8.314 * 300.15) = 9.86 moles.

Assuming that the gas is diatomic (like nitrogen or oxygen), the heat capacity (C) is 7/2R = 7/2 * 8.314 = 29.1 J/(mol*K).

So, ΔH = Q = nCΔT = 9.86 * 29.1 * (400.15 - 300.15) = 28600 J

Finally, we can substitute ΔH, Tf, and ΔS into the Gibbs-Helmholtz equation to find ΔG:

ΔG = ΔH - TfΔS = 28600 J - 400.15K * 1 J/K = 24600 J

So, the change in Gibbs free energy is 24600 Joules.