# Video: Calculating the Thermodynamic State Parameters in an Adiabatic Quasi-Static Process

An ideal monatomic gas at a temperature of 542 K expands adiabatically and reversibly to 4.00 times its volume. What is the gas’s final temperature?

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### Video Transcript

An ideal monatomic gas at a temperature of 542 kelvin expands adiabatically and reversibly to 4.00 times its volume. What is the gas’s final temperature?

Considering this problem statement, one of the key things we’re told is that when the gas expands, it does so adiabatically, that is, without exchanging heat with the system. In the case of an expansion like this, we can say that the temperature of the gas multiplied by the volume raised to the power of 𝛾 minus one — we’ll explain what 𝛾 is in a second — is equal to a constant.

This means that if we have an initial state, that is, before the expansion, and a final state, after the expansion, we can write that 𝑇 sub 𝑖 times 𝑉 sub 𝑖 to the 𝛾 minus one is equal to 𝑇 sub 𝑓 times 𝑉 sub 𝑓 to the 𝛾 minus one.

Now about this factor 𝛾, this factor 𝛾 is equal to a ratio: the specific heat of the gas at a constant pressure, 𝑃, to the specific heat of the gas at a constant volume, 𝑉. The fact that our ideal gas is monatomic in this example implies a particular value for 𝛾. It implies that the ratio 𝐶 sub 𝑃 to 𝐶 sub 𝑉 is equal to five-thirds.

Going back to our expression for temperature then, we want to solve for the final temperature of the system, 𝑇 sub 𝑓. Rearranging, that’s equal to 𝑇 sub 𝑖 times the ratio 𝑉 sub 𝑖 over 𝑉 sub 𝑓, all raised to the 𝛾 minus one power.

In the problem statement, we’re told that 𝑇 sub 𝑖 is 542 kelvin and that 𝑉 sub 𝑓 is four times greater than 𝑉 sub 𝑖. In other words, 𝑉 sub 𝑖 over 𝑉 sub 𝑓 is one-quarter. Since we also know the value for 𝛾, we’re ready to plug in and solve for 𝑇 sub 𝑓. Five-thirds minus one is two-thirds. And when we plug this expression into our calculator, to three significant figures, we find that 𝑇 sub 𝑓 is 215 kelvin. That’s the final temperature of the gas.