Question

If $\mathbf{54}\mathbf{.0}\mathbf{}\mathbf{\text{g}}$argon at$\mathbf{\text{400K}}$ is compressed isothermally and reversibly from a pressure of$\mathbf{\text{1.50}}$ to$\mathbf{4.00}\mathit{a}\mathit{t}\mathit{m}$ , calculate the work done on the gas and the heat absorbed by gas in the process. What are the changes in energy $\mathbf{\left(}\mathit{\Delta }\mathit{U}\mathbf{\right)}$and in enthalpy$\mathbf{\left(}\mathit{\Delta }\mathit{H}\mathbf{\right)}$ of the gas?

Short answer

Change in energy,$\Delta U=0$.

Change in enthalpy, $\Delta H=0$.

Step-by-step solution

Given data

If $4.00atm$ argon at $400\text{K}$ is compressed isothermally and reversibly from a pressure of $1.50$ to $4.00atm$.

Concept of Isothermal process

The vapor Isothermal processes occur under constant temperature.

$\mathit{\Delta }\mathit{T}\mathbf{=}\mathbf{0}$

The constant temperature is acquired by exchanging the heat with the surroundings.

When talking about ideal gases, the internal energy change is only dependent on its temperature (Joule's second law). This is because the ideal gas has no intermolecular forces (that is the assumption).

Therefore, if the temperature is held constant, the change in internal energy is constant.

The same thing goes for the enthalpy change.

Concept the number of moles of argon

Amount of argon, $m=54.0\text{g}$.

Molar mass of argon, $M=39.95\text{g}/\text{mol}$.

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{54.0}{39.95}\\ n=1.35\text{mo}\end{array}$

Calculate the work done on the gas

The formula to calculate the work done on the gas is:

$\begin{array}{c}\mathrm{w}=-\mathrm{nRTln}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\\ \mathrm{w}=-1.35\times 8.314\times 400\times \ln \frac{1.5}{4.0}\\ \mathrm{w}=4403.5\text{J}\end{array}$

The reason for the sign being negative is because the work done by the surroundings (which is the same in value but different in sign as the work done by the system).

Calculate the heat absorbed by the gas

The heat absorbed by the gas is calculated by the formula for the change in internal energy.

$\Delta U=w+q$ ....... (1)

Set up a relation between the heat and the work done on the gas:

Substitute the values in equation (1)

$\begin{array}{c}\Delta U=w+q\\ =-4403.5J+4403.5J\\ =0\end{array}$

Calculate the change in enthalpy

The formula for the change in enthalpy for an isothermic process is,$\Delta H=\Delta U+\Delta \left(pV\right)$.

Since the change in internal energy is 0 for isothermal processes on ideal gases is 0.

$\begin{array}{c}\Delta H=\Delta U+\Delta \left(pV\right)\\ =0\end{array}$