# Why am I getting a different answer than my book? [closed]

Problem:

$$1 mol$$ helium gas is kept at $$2\times10^5Pa$$ pressure and $$600K$$ temperature in a heat-insulating cylinder on top of which a frictionless piston is attached. Then the cylinder's pressure was lessened to half of the original pressure. Determine the final temperature of the gas in the insulated cylinder.

My book's solution:

Here,

Initial pressure, $$P_1=2\times10^5Pa$$

Initial temperature, $$T_1=600K$$

Final pressure, $$P_2=10^5Pa$$

$$\gamma=1.67$$

Final temperature, $$T_2=?$$

$$T_1P_1^{\frac{1-\gamma}{\gamma}}=T_2P_2^{\frac{1-\gamma}{\gamma}}$$

$$T_2={\left(\frac{P_1}{P_2}\right)}^{\frac{1-\gamma}{\gamma}}\times T_1$$

$$T_2={\left(\frac{2\times10^5}{10^5}\right)}^{\frac{1-1.67}{1.67}}\times 600$$

$$T_2=454.33766K$$

My attempt:

Here,

Initial pressure, $$P_1=2\times10^5Pa$$

Initial temperature, $$T_1=600K$$

Final pressure, $$P_2=10^5Pa$$

Final temperature, $$T_2=?$$

Applying Charles' law, we get,

$$\frac{P_1}{T_1}=\frac{P_2}{T_2}$$

$$T_2=300K$$

Why doesn't Charles' law work here?

• The volume has changed. Dec 3, 2021 at 19:59
• Correction: it will be Gay Lussac's law of pressure instead of Charles' law. Dec 3, 2021 at 20:18

By setting $$P_1/T_1 = P_2/T_2$$, you are assuming that the volume of the gas is constant, c.f. the ideal gas law: $$\frac{P}{T} = \frac{N k_B}{V}$$

In this case, the pressure, volume, and temperature are changing, so one has only that $$P_1 V_1/T_1 = P_2 V_2/T_2$$. This could be re-expressed as $$\frac{T_2}{T_1} = \frac{P_2}{P_1}\frac{V_2}{V_1}$$ Since you only know $$P_2/P_1$$, you need more information to find $$T_2/T_1$$. That information comes in the statement that the cylinder is insulating, which means that the process in question is adiabatic.

Charles' "law" must only hold for fixed volumes. Starting from the ideal gas law: $$PV=NkT$$, $$\frac{P}{T}=\frac{Nk}{V}$$ so the ratio $$P/T$$ is only fixed if $$V$$ is held fixed. Here, you have a "frictionless piston" that is allowed to move and change the volume of the cylinder. However, you do know that the cylinder is thermally insulated, so $$Q=0$$. The formula your book uses is derived from combining the first law of thermodynamics $$\Delta U = Q+W$$ with the equipartition theorem $$U=\frac{f}{2} NkT$$, and the ideal gas law $$PV = NkT$$ with the assumption that $$Q=0$$.

Charles' law is for a constant pressure process, not a constant volume process. In any case, you are dealing with an adiabatic (Q=0) expansion which is neither constant pressure nor constant volume.

That said, the book answer is correct if one can assume that the adiabatic process is reversible. It should have stated so, because without that assumption you wouldn't have enough information to determine the final temperature as pointed out by @J.Murray.

Hope this helps.

• @tryingtobeastoic OK, but that law applies to constant volume process which this is not. Dec 3, 2021 at 20:19