Video: Work, Heat, and Energy Calculations Using the First Law of Thermodynamics in an Isothermal Process

An ideal gas expands quasi-statically and isothermally from a state with pressure 2.2 bar and volume 0.85 L to a state with volume 1.70 L. How much heat is added to the expanding gas?

03:35

Video Transcript

An ideal gas expands quasi-statically and isothermally from a state with pressure 2.2 bar and volume 0.85 liters to a state with volume 1.70 liters. How much heat is added to the expanding gas?

We can call this heat added to the expanding gas 𝑄 and start solving for it by recalling the first law of thermodynamics. This law tells us that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

If we were to plot this expansion on a pressure versus volume chart, we’re told the gas starts at a pressure of 2.2 bars and a volume of 0.85 liters. And we’re told it ends up with a volume of 1.70 liters though we don’t know it’s final pressure. We do know though that this gas expansion happens isothermally. So that if we draw an isotherm of constant temperature, we know that the expansion happens along this line.

This then is the path that the gas follows as it expands. And we want to solve for how much heat is added to the gas during this expansion. We’re helped in solving for this heat by the fact that our ideal gas expands isothermally which implies that Δ𝑒 is equal to zero. And this implies that 𝑄, the heat added to the explaining gas, equals π‘Š, the work done by the gas. So if we can solve for work, we’ll have effectively solved for heat, 𝑄.

Knowing that, we can recall that the work done by a gas is equal to the area under the curve of its 𝑃𝑉 plot. If we integrate to solve for this area under the curve, that will tell us the work done by the gas which will tell us how much heat is added. We can write, as an equation, that the work done, the area under this curve, is equal to the integral of the pressure times 𝑑𝑉 from the initial volume to the final volume, 𝑉 one to 𝑉 two.

This integral though is a bit deceptive in its simplicity because 𝑃 is actually a function of the volume 𝑉. The ideal gas law tells us that 𝑃 times 𝑉 is equal to 𝑛𝑅𝑇 or that 𝑃 is equal to 𝑛𝑅𝑇 divided by volume 𝑉. So we substitute this expression in for the pressure 𝑉.

And now we have an integral in terms of 𝑑𝑉 over 𝑉. The factor 𝑛 times 𝑅 times 𝑇 doesn’t depend on volume so we pull it out of the integral. And when we integrate, we find a result of 𝑛 times 𝑅 times 𝑇 times the natural log of 𝑉 two over 𝑉 one. The only trouble with this result is that we don’t know 𝑛, the number of moles of the gas, or 𝑇, its temperature. Once again, referring to the ideal gas law, we see that 𝑃 times 𝑉 is equal to 𝑛𝑅𝑇. So we can substitute 𝑃𝑉 in for that expression.

We now have an expression for the work done by the gas in terms of known values. And since we can solve for the work, we can solve for the heat added. For the value of 𝑃, pressure, we’ll use our given value of 2.2 bars. And for our volume, we’ll use our initial system volume of 0.85 liters. Before we calculate our final result for the work done by the gas, and therefore the heat added to it, we’ll want to do a bit of unit converting.

Let’s change the pressure from units of bars to units of pascals. And we’ll change the volume from units of liters to units of cubic meters. We multiply these terms by the appropriate conversion factors, so we now have pressure in pascals and volume in cubic meters, and enter this whole expression on our calculator. To two significant figures, we find that 𝑄 is 130 joules. That’s how much heat is added to the expanding gas.

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