Question

One mole of an ideal gas with a volume of 1.0 L and apressure of 5.0 atm is allowed to expand isothermally into an evacuated bulb to give a total volume of 2.0 L. Calculate w and q. Also calculate ${\mathit{q}}_{\mathbf{r}\mathbf{e}\mathbf{v}}$ for this change of state.

Short answer

The value of w and q are 0and${\mathit{q}}_{\mathbf{r}\mathbf{e}\mathbf{v}}$ for this change of state is $\mathbf{350}\mathbf{.}\mathbf{95}\mathit{J}$.

Step-by-step solution

Step 1: Introduction to the Concept

Internal energy of the system is defined as the total of the kinetic and potential energy of the particles in the system.

Change in internal energy is expressed mathematically as:

$\mathit{\Delta }\mathit{E}\mathbf{=}\mathit{q}\mathbf{+}\mathit{w}$

WhereE denotesinternal energy,q denotesheat, and w denoteswork performed by or on the system.

Step 2: Determination of the value of q and w

From the given,

For 1 mole of ideal gas:

Volume$=1.0\text{L}$

Pressure$=5.0\text{atm}$

In an evacuated bulb, expand isothermally to yield a total volume$=2.0\text{L}$

Because the internal energy changes in an isothermal process is constant.

In the case of an isothermal process,$\Delta E=0$for an ideal gas.

$0=q+w$

$q=-w$

Free expansion is a technique in which gas expands isothermally into an evacuated bulb.As a result, no work is done because no pressure is applied.

$q=-w$

Put $w=0andq=0$in the equation,

As a result, the value of q for an isothermal process is zero.

Step 3: Determination of the heat evolved in a reversible isothermal process

The change in internal energy for an isothermal process is zero.

Therefore, $q=-w$

We are going to use the following relation to calculate the work done in a reversible isothermal process:

$\mathbf{w}\mathbf{=}\mathbf{-}\mathbf{nRTln}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{V}}_{\mathbf{1}}}$

$$\begin{gathered} {\text{V}}_{\text{1}}=1.0 \text{L} \\ {\text{V}}_2=2.0 \text{L} \end{gathered}$$

$$\begin{gathered} w=-nRTln\frac{V_2}{V_1} \\ =-1.00\times 8.314J/molK\times 60.9K\times ln\frac{2}{1} \\ =-350.95J \end{gathered}$$

Therefore, the heat evolved in a reversible isothermal process is$350.95 \text{J}$.