Question

1 mole of ideal monoatomic gas is heated by supplying 5 kJ heat from a reservoir maintained at 400 K from 300 K to 400 K. In the process volume of gas increased from 1 L to 10 L. Find $\Delta S_{total}$ (in J/K-mol) in the process

Use : $\ln\left(\frac{4}{3}\right)=0.3$, $\ln 10=2.3$, R=8.3 J/K-mol and $\ln x=2.3 \log x$

Answer 1

10.325

Answer 2

To find the total entropy change ($\Delta S_{total}$), we need to calculate the entropy change of the system ($\Delta S_{system}$) and the entropy change of the surroundings ($\Delta S_{surroundings}$).

1. Calculate $\Delta S_{system}$: For an ideal gas, the entropy change is given by the formula: $\Delta S_{system} = n C_V \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right)$ Given:

• Number of moles, $n = 1$ mol
• Initial temperature, $T_1 = 300$ K
• Final temperature, $T_2 = 400$ K
• Initial volume, $V_1 = 1$ L
• Final volume, $V_2 = 10$ L
• Gas constant, $R = 8.3$ J/K-mol
• For a monoatomic ideal gas, $C_V = \frac{3}{2}R$.

Substitute the values into the formula: $\Delta S_{system} = (1 \text{ mol}) \times \left(\frac{3}{2}R\right) \times \ln\left(\frac{400 \text{ K}}{300 \text{ K}}\right) + (1 \text{ mol}) \times R \times \ln\left(\frac{10 \text{ L}}{1 \text{ L}}\right)$ $\Delta S_{system} = \frac{3}{2}R \ln\left(\frac{4}{3}\right) + R \ln(10)$ Using the given values: $\ln\left(\frac{4}{3}\right)=0.3$ and $\ln 10=2.3$: $\Delta S_{system} = 1.5 \times (8.3 \text{ J/K-mol}) \times (0.3) + (8.3 \text{ J/K-mol}) \times (2.3)$ $\Delta S_{system} = (12.45 \times 0.3) \text{ J/K-mol} + (19.09) \text{ J/K-mol}$ $\Delta S_{system} = 3.735 \text{ J/K-mol} + 19.09 \text{ J/K-mol}$ $\Delta S_{system} = 22.825 \text{ J/K-mol}$

2. Calculate $\Delta S_{surroundings}$: The heat is supplied from a reservoir maintained at a constant temperature. For heat transfer from a reservoir, the entropy change of the surroundings is given by: $\Delta S_{surroundings} = \frac{q_{surroundings}}{T_{reservoir}}$ The heat supplied to the system ($q_{system}$) is 5 kJ = 5000 J. Therefore, the heat lost by the surroundings ($q_{surroundings}$) is -5000 J. The temperature of the reservoir ($T_{reservoir}$) is 400 K. $\Delta S_{surroundings} = \frac{-5000 \text{ J}}{400 \text{ K}}$ $\Delta S_{surroundings} = -12.5 \text{ J/K}$

3. Calculate $\Delta S_{total}$: The total entropy change is the sum of the entropy change of the system and the entropy change of the surroundings: $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}$ $\Delta S_{total} = 22.825 \text{ J/K-mol} + (-12.5 \text{ J/K})$ $\Delta S_{total} = 10.325 \text{ J/K-mol}$

The positive value of $\Delta S_{total}$ indicates that the process is irreversible and spontaneous.