The question is as follows:

A sample of $\pu{4.0 moles}$ of a gas ($C_{v,m} = \pu{21 J mol-1 K-1}$) has an initial pressure of $\pu{304.4 kPa}$ and an initial volume of $\pu{20 dm3}$ at $\pu{270 K}$. It undergoes an adiabatic expansion against a constant external pressure of $\pu{79.8 kPa}$, until the volume has increased by a factor of $3$. Calculate the final temperature.

I have calculated: $Q = 0$ and $w = \pu{-3192 J}$ and change in internal energy $=\Delta U = \pu{-31922 J}$.

The next question is to calculate the final temperature of the gas. However, I am unsure on how to calculate this. Originally I thought as $Q=0$, then the final temperature would be the same as the initial temperature. However, I have realized this is incorrect.

  • 1
    $\begingroup$ What is the relationship between change in internal energy and change in temperature for an ideal gas? $\endgroup$ – Chester Miller Mar 16 '17 at 2:05

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic.

In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then its temperature will not remain constant during any expansion or compression.

In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.

The heat change is zero then $dq=0$ which means from the First Law $$dU = dw$$ and so $$ C_vdT = -pdV$$

Dividing both sides by T and for one mole of an perfect gas $p=RT/V$ thus $$C_v\frac{dT}{T}=-R\frac{dV}{V}$$ If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give $$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$ or $$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$

using the relationship $C_p = C_v+R$ $$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$ Using the gas law this van be rewritten in a more useful form as $$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$ where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.

The change in internal energy in an adiabatic process is $$\Delta U=C_v(T_2-T_1)$$

Adiabatic reversible

In a reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.

The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$ and as only $V_2$ is unknown this can be calculated. The work done is $$w=nC_v(T_2-T_1)$$

Adiabatic irreversible

In an irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume.
The internal energy change is $\Delta U= nC_v(T_2-T_1)$ and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained. Similarly if the volumes are known the final temperature can be obtained.

Isothermal irreversible

In an isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is $$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$ thus expansion into a vacuum does no work.

Isothermal reversible

A reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas, $$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$

$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$


Work done in adiabatic process $W =\frac {nR \Delta T}{1-\gamma}$

$Cv$ (specific heat at constant volume) $= 21 \frac{\mathrm{J}}{\mathrm{mol}*\mathrm{K}}$

$Cv = \frac{fR}{2} \to f = \frac{21\times2}{8.314}= 5$ So $\gamma= \frac{f+2}{f}$, where $f$ is the degree of freedom.

$\to γ = \frac7 5$

Now work done at constant pressure $= 79.8\times(1000)\times[60(10^{-3}) - 20(10^{13})] = 3192 \,\mathrm{J}$ (positive work done by gas. The internal pressure will simultaneously be equal to the external pressure otherwise first law of thermodynamics won't hold true.)

$W = \frac{nR ΔT}{1-γ}$

$3192 = \frac{4(8.316)(T - 270)}{1-\frac{7}{5}}$

$T = [\frac{3192(-0.4)}{4(8.316)}] + 270$

$\to T = 231.61 K$


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