Reversible Expansion of an Ideal Gas under Isothermal and also Adiabatic Conditions (First Law of Thermodynamics) (Example 1)

One mole of an ideal gas with $${C_{v,m}} = \frac{3}{2}R$$ expands reversibly from 24.6 L and 300 K to 49.2 L. Find the final temperature and pressure for when the expansion is a.) isothermal b.) adiabatic

Let's consider isothermal gas expansion first. If it's isothermal, then T1=T2 so T2=300K. As well, for isothermal expansion, $$\Delta U = 0$$, $$\Delta H = 0$$, and so $$Q = - W$$.

You can just use the ideal gas law,

${P_2} = \frac{{nR{T_2}}}{{{V_2}}}$

Substitute in the given values.

${P_2} = \frac{{(1mol)\left( {0.08206\frac{{Latm}}{{molK}}} \right)(300K)}}{{49.2L}}$

${P_2} = 0.5atm$

Done, we found the final temperature and pressure for the case of isothermal.

Now for the second part, let's find the final temperature and pressure for the case of adiabatic gas expansion. Adiabatic just means that Q=0, so $$\Delta U = W$$.

We're given in the question that $${C_{v,m}} = \frac{3}{2}R$$ which is usually the assumption for ideal gases. To get $${C_{p,m}}$$, we add 1R to $${C_{v,m}}$$, this is by definition. So $${C_{p,m}} = \frac{5}{2}R$$.

Now we can find $$\gamma$$. By definition it is:

$\gamma = \frac{{{C_{p,m}}}}{{{C_{v,m}}}}$

$\gamma = \frac{{\frac{5}{2}R}}{{\frac{3}{2}R}}$

$\gamma = 1.6667$

And for adiabatic gas expansion we can use:

$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}$

Substitute in all known values and we can solve for T2.

$\frac{{{T_2}}}{{300K}} = {\left( {\frac{{24.6L}}{{49.2L}}} \right)^{1.6667 - 1}}$

${T_2} = 188.99K$

And now substitute in this into the ideal gas law:

${P_2} = \frac{{nR{T_2}}}{{{V_2}}}$

${P_2} = \frac{{(1mol)\left( {0.08206\frac{{Latm}}{{molK}}} \right)(188.99K)}}{{49.2L}}$

${P_2} = 0.32atm$

Done, we found the final temperature and pressure for the case of adiabatic gas expansion. It's worth memorizing these cases and equations.