Question

Suppose 4.00mol of an ideal gas undergoes a reversible isothermal expansion from volume${\mathit{v}}_{\mathbf{1}}$ to${\mathit{v}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{V}}_{\mathbf{1}}$ volume at temperature T = 400 K . (a) Find the work done by the gas and (b) Find the entropy change of the gas (c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?

Short answer

1. The work done by the gas is $9.22\times {10}^3\mathrm{J}$
2. The entropy change of the gas is$23.1\mathrm{J}/\mathrm{K}$
3. If expansion is reversible and adiabatic instead of isothermal, the entropy change of the gas is zero.

Step-by-step solution

The given data

1. $V_2=2.00V_1$Mole of an idea gas,$\mathrm{n}=4.00\mathrm{mol}$
2. Volume change in the isothermal process, from $V_1$to
3. Change takes place at temperature, T = 400 K

Understanding the concept of isothermal and adiabatic expansion change

In an isothermal process, the temperature stays constant; however, in an adiabatic process, the temperature changes because internal energy is changed rather than heat being transported. We can find the work done by the gas using the formula for work done in an isothermal process. Then the entropy change of the gas can be found using its formula for the isothermal process.

Formulae:
The work done by the gas during isothermal process, $W=nRTIn\left(\frac{V_2}{V_1}\right)$ …(i)

Where, R is gas constant = $8.314\mathrm{J}.{\mathrm{mol}}^{-1}.{\mathrm{K}}^{-1}$

The entropy change by the gas, $∆S=\frac{Q}{T}$ …(ii)

(a) Calculation of the work done by the gas

During the isothermal process, work done by the gas using equation (i) and the given values can be given as:

$$\begin{gathered} W=\left(4\mathrm{mol}\right)\left(8.314\mathrm{J}.{\mathrm{mol}}^{-1}.\mathrm{K}\right)\left(400\mathrm{K}\right)In\left(\frac{2V_1}{V_1}\right) \\ =9220\mathrm{J} \\ \approx 9.22\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the work done by the gas is$9.22\times {10}^3\mathrm{J}$

(b) Calculation of the entropy change by the gas

In isothermal process, the heat transferred b y the body is equal to the work done by the gas for constant temperature. So, Q = W

From equation (ii) and using the given values, we can get the entropy change by the gas as follows:

$$\begin{gathered} ∆S=\frac{W}{T} \\ =\frac{9220\mathrm{J}}{400\mathrm{K}} \\ =23.05\mathrm{J}/\mathrm{K} \\ \approx 23.1\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change by the gas is 23.1 J/K

(c) Calculation of the entropy change by the gas in adiabatic process

For reversible adiabatic processes, entropy is constant. Hence, entropy change of the gas, if the expansion is reversible and adiabatic is zero.