Reversible Expansion of an Ideal Gas under Isochloric (Same Volume) and Isobaric (Same Pressure), Find Q,W,U,H (First Law of Thermodynamics) (Example 2)

The last few examples earlier were isothermic and adiabatic. The other two kinds of problems are isochloric and isobaric.

2 g of helium gas has \({C_{p,m}} = \frac{5}{2}R\). Assume it is an ideal gas. Find Q, W, \(\Delta H\), \(\Delta U\), for

a.) Reversible constant-pressure expansion from 20L to 40L at 0.8 bar (isobaric expansion).

b.) Reversible heating, where the pressure increases from 0.6 to 0.9 bar, while volume remains constant at 15 L (isochloric expansion).

Let's do a.) first, the isobaric case. To get Cvm from Cpm, subtract 1R. And find gamma as the ratio.

\[{C_{p,m}} = \frac{5}{2}R\]

\[{C_{v,m}} = \frac{3}{2}R\]

\[\gamma = \frac{{{C_{p,m}}}}{{{C_{v,m}}}} = \frac{{\frac{5}{2}R}}{{\frac{3}{2}R}} = \frac{5}{3}\]

We can use the ideal gas law to find T1. Note that 0.8 bar is roughly 0.789 atm.

\[{T_1} = \frac{{{P_1}{V_1}}}{{nR}}\]

\[{T_1} = \frac{{(0.789atm)(20L)}}{{\left( {\frac{{2gHe}}{{4.002g/molHe}}} \right)\left( {0.08206\frac{{Latm}}{{molK}}} \right)}}\]

\[{T_1} = 385.05K\]

A trick here is that since it's constant pressure, V and T are directly proportional. So T2 = 2T1 = 770.10 K.

Now we can find Q, W, \(\Delta H\), and \(\Delta U\). Note that R is a "different" value because of different units!

\[W = - nR\Delta T\]

\[W = - \left( {\frac{{2gHe}}{{4.002g/molHe}}} \right)\left( {8.314\frac{J}{{molK}}} \right)(770.10K - 385.05K)\]

\[W = - 1599.87J\]

\[\Delta U = n{C_v}\Delta T = \left( {\frac{{2g}}{{4.002g/mol}}} \right)\left( {\frac{3}{2}} \right)\left( {8.314\frac{J}{{molK}}} \right)(770.10K - 385.05K)\]

\[\Delta U = 2399.78J\]

\[\Delta H = n{C_P}\Delta T = \left( {\frac{{2g}}{{4.002g/mol}}} \right)\left( {\frac{5}{2}} \right)\left( {8.314\frac{J}{{molK}}} \right)(770.10K - 385.05K)\]

\[\Delta H = 4000J\]

\[q = \Delta U - W = 2400J - ( - 1600)\]

\[q = 4000J\]

That's the isobaric case, all information calculated! Now let's do part b, which is the constant volume (isochloric) case.

For a constant volume process, W = 0.

We can find T1 from the ideal gas law. Note that we convert 0.6 bar to 0.592 atm.

\[{T_1} = \frac{{{P_1}{V_1}}}{{nR}}\]

\[{T_1} = \frac{{(0.592atm)(15L)}}{{\left( {\frac{{2gHe}}{{4.002g/molHe}}} \right)\left( {0.08206\frac{{Latm}}{{molK}}} \right)}}\]

\[{T_1} = 216.59K\]

The V is constant, so T and P have the following relation:

\[{T_2} = {T_1}\left( {\frac{{{P_2}}}{{{P_1}}}} \right)\]

\[{T_2} = (216.59K)(1.5)\]

\[{T_2} = 324.88K\]

\[\Delta U = n{C_v}\Delta T = \left( {\frac{{2g}}{{4.002g/mol}}} \right)\left( {\frac{3}{2}} \right)\left( {8.314\frac{J}{{molK}}} \right)(324.88K - 216.59K)\]

\[\Delta U = 674.94J\]

\[\Delta H = n{C_P}\Delta T = \left( {\frac{{2g}}{{4.002g/mol}}} \right)\left( {\frac{5}{2}} \right)\left( {8.314\frac{J}{{molK}}} \right)(324.88K - 216.59K)\]

\[\Delta H = 1124.89J\]

For isochloric expansion:

\[Q = \Delta U - W = \Delta U\]

\[Q = 674.99J\]

Done, we solved for all variables using the isochloric expansion case. Understanding adiabatic, isothermal, isochloric, and isobaric expansion examples pretty much means you know how to solve first law thermodynamics problems!

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