91Ó°ÊÓ

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

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, $W=nRTIn\left(\frac{V_2}{V_1}\right)$ …(¾±)

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

The entropy change by the gas, $∆S=\frac{Q}{T}$ …(¾±¾±)

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

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

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

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:

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

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

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