Question

Suppose$\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\text{mol}}$of a monatomic ideal gas$\left(c_V=\frac{3}{2}R\right)$ is expanded adiabatically and reversibly from a temperature$\mathit{T}\mathbf{=}\mathbf{300}\mathbf{}\mathit{K}$ , where the volume of the system is $\mathbf{20}\mathbf{.0}\mathbf{}\mathbf{\text{L}}$, to a volume of$\mathbf{60}\mathbf{.0}\mathbf{}\mathbf{\text{L}}$ . Calculate the final temperature of the gas, the work done on the gas, and the energy and enthalpy changes.

Short answer

The final temperature of the gas is $144.22\text{K}$and work done is $3890\text{J}$.

The energy change is $3890\text{J}$ and enthalpy change is$6485\text{J}$ .

Step-by-step solution

Given information

The observed adiabatic and reversible expansion of a monoatomic ideal gas where the number of moles, temperature and volume is known:

$n=2.00mol{T}_1=300K{V}_1=20.00L{V}_2=60.00L$.

Concept of adiabatic processes

Adiabatic expansion is defined as the optimal behaviour for a closed system with constant pressure and decreasing temperature. Adiabatic processes occur without any heat exchange.

Calculate the final temperature

Work done by the ideal gas when it expands $dw=-pdV$.

The adiabatic change means no heat transfer therefore the change in internal energy for an ideal gas is:

$\begin{array}{c}\mathrm{dU}=\mathrm{dw}\\ ={\mathrm{c}}_{\mathrm{v}}\mathrm{dT}\end{array}$

This means that $-\mathrm{pdV}={\mathrm{c}}_{\mathrm{v}}\mathrm{dT}$.

Let’s combine with that with the ideal gas equation $pV=nRT$.

It is obtained that $\frac{-pdV}{T}=\frac{vdT}{V}$.

The expression is $c_{\mathit{v}}ln\frac{{\mathit{T}}_{\mathit{2}}}{{\mathit{T}}_{\mathit{1}}}\mathit{=}\mathit{-}nR\frac{{\mathit{V}}_{\mathit{2}}}{{\mathit{V}}_{\mathit{1}}}$.

By rearranging the expression ,it is obtained $\frac{c_v}{nR}\times ln\frac{T_2}{T_1}=ln\frac{V_1}{V_2}$.

Let's substitute the first factor with $x=\frac{c_v}{nR}$.

The equation obtained is $ln{\left(\frac{{\mathrm{T}}_{\mathrm{2}}}{{\mathrm{T}}_{\mathrm{1}}}\right)}^x=ln\frac{V_1}{V_2}$.

Or: ${\left(\frac{{\mathrm{T}}_{\mathrm{2}}}{{\mathrm{T}}_{\mathrm{1}}}\right)}^x=\frac{V_1}{V_2}$

So, final formula is $T_2={\left(\frac{{\mathrm{V}}_{\mathrm{1}}}{{\mathrm{V}}_{\mathrm{2}}}\right)}^{x}qq$.

Calculate the value of  x

First find the value of $x$to put into the final formula.

data-custom-editor="chemistry" $\begin{array}{l}x=\frac{c_v}{nR}\\ x=\frac{\frac{3}{2}R}{R}\\ x=1.5\end{array}$

Then substitute to calculate the final temperature.

data-custom-editor="chemistry" $\begin{array}{l}{T}_2=300\times {\left(\frac{20.0}{60.0}\right)}^{\frac{1}{1.5}}\\ T_2=144.22K\end{array}$

Calculate the work done

The work done on the gas as per formula $w=-n{c}_v\Delta T$.

The reason for the sign being negative is because we are looking for the work done on the gas. Calculate the work done on the gas as:

$\begin{array}{c}w=-n{c}_v\Delta T\\ w=2.00\times 12.47\times (144-300)\\ w=3890J\end{array}$

Calculate internal thermal energy change

The usual formula $\Delta U=w+q$.

Since the heat transferred for adiabatic processes on ideal gases is 0, set up a relation between the internal energy and the work done on the gas as $\Delta U=w$.

Therefore, the change in internal energy is $\Delta U=3890J$.

Calculate change in enthalpy for adiabatic process

The formula for change in enthalpy for an adiabatic process is$\Delta H=-n{c}_p\Delta T$.

The value of heat capacity, the rule is $c_p=c_v+R$.

Therefore calculate:

$\begin{array}{c}{c}_p=\frac{3}{2}R+R\\ =\frac{5}{2}R\\ c_p=20.78J/molK\end{array}$

Calculate the enthalpy change

Let’s calculate the enthalpy change, put everything in the formula:

$\begin{array}{c}\Delta H=-2.00\times 20.78\times (144-300)\\ \Delta H=6485J\end{array}$