Example 4B.1

[105743571]

Example 4B.1
Engineers designing new piston engines and turbines need to understand how work and heat are involved in various compression and expansion cycles. Suppose that 1.00 mol of ideal gas molecules at an initial pressure of 3.00 atm and 292 K expands against a constant external pressure of 0.20 atm from 8.00 L to 20.00 L by two different paths. (a) Path A is an isothermal, reversible expansion.

For Path A, since it's isothermal, this means that q = o right? And, you would only need to calculate the energy lost by work (nRTln(V2/V1). Then why does the textbook say that q = -w?

[Shria G 2D]

Hi,
I don't think q is always 0 for isothermal expansion. If it says reversible, isothermal expansion of an ideal gas, I think we just know that the change in internal energy (delta U) is 0 and isothermal means the temperature is constant. But, just because the temperature is constant, that doesn't mean energy can't be transferred as heat to or from the system. Also, the textbook says that delta U = 0 because when an ideal gas expands isothermally, its molecules are moving at the same speed so the potential and kinetic energy stay the same which is why the internal energy does not change. Since delta U = q + w and delta U = 0 in this case, q + w = 0 so q = -w. I hope that helps!

[Rena Wu 3E]

Hello, since the question states that Path A is an isothermal, reversible expansion, we know that the change in internal energy of the system (deltaU) is equal to q + w. But because the system is isothermal, we also know that the value of deltaU is equal to 0. Putting these two facts together, we get q + w = 0. The textbook says q = -w since they just subtracted w on both sides. Basically, isothermal wouldn't necessarily mean that q = 0, since there can still be a transfer of energy in a system without a change in temperature (e.g., phase changes). Hope this helps!