Answer to Question #329460 in Molecular Physics | Thermodynamics for Quân Jason

Question

What is delta S, the change in entropy of ten moles of an ideal monatomic gas that is expanded slowly and isothermally at a temperature of 293K from a pressure of 2 atm to a pressure of 1 atm?

a) 57.6 J/K

b) 203.4 J/K

c) 21.7 J/K

Answer 1

Change of entropy is given as follows:

$$\Delta S = \dfrac{\delta Q}{T}$$

where $\delta Q$ is the heat supplied to the system, and $T=293K$ is the temperature of the process.

According to the first law of thermodynamics, the heat supplied to the system is:

$$Q = -W$$

where $W$ is the work done on the system by its surrounding. Since the temperature remains constant, the internal energy of the system does not change.

The work in isothermal process is given as follows (see https://en.wikipedia.org/wiki/Isothermal_process#Calculation_of_work):

$$W = -nRT\ln\left( \dfrac{V_2}{V_1} \right)$$

where $n=10mols$ is the amount of gas, $V_2, V_1$ are the final and initial volumes respectively.

According to the Boyle's Law, these volumes are connected with the corresponding pressures as follows:

$$\dfrac{V_2}{V_1} = \dfrac{p_1}{p_2}$$

Finally, obtain:

$$\Delta S = \dfrac{1}{T}nRT\ln\left( \dfrac{V_2}{V_1} \right) = nR\ln\left( \dfrac{p_1}{p_2} \right)\\ \Delta S=10mole\cdot 8.3\dfrac{J}{K\cdot mol}\cdot \ln\left( \dfrac{2atm}{1atm} \right) \approx 57.5 J/K$$

Answer. a) 57.6 J/K.