# Adiabatic volume change of an ideal gas thought process

I am struggling with trying to solve the following problem, attempting just (a) and the first part of (b) for this exercise:

During a reversible adiabatic volume change of an ideal gas, $PV^{\gamma}$ remains constant, where $\gamma$ is the ration of the heat capacity at constant volume to that at constant pressure. i.e. $\gamma = C_P/C_V$. One mole of an ideal gas, initially at $300^{\circ}K$ and atmospheric pressure, is compressed to half it's initial volume.

(a) Derive a general expression for the work done on the gas in terms of the starting and finishing volumes and other necessary variables. Remember that you'll need to find an expression for $P$ as a function of $V$, and that this expression will involve $P_1$ and $V_1$.

(b) Use yor expression to calculate the amount of work done for a monatomic ideal gas and (b) a diatomic gas, in which the vibrational motion is completely frozen out.

Here's my initial thought process:

• This is an adiabatic process, so $PV^{\gamma} = K$ where $K$ is a constant.
• This gas is monatomic, so $\gamma = 1 + 2/3$.
• We are not given a volume, but given a pressure and temperature, which means the volume can be deduced from the ideal gas law.
• This is an adiabatic volume change, so the work can be found given the following integral:

$$W = - \int_{V_1}^{V_2} PdV$$

• Since $V_2 = 1/2 V_1$, we can replace that for $V_2$, and note $P=\frac{K}{V^\gamma}$. Thus we want to solve:

$$W = - K\int_{V_1}^{0.5V_1} \frac{dV}{V^\gamma}$$

This should reduce to the following equation:

$$W = -K \left(\frac{0.5V_1^{1-\gamma} - V_1^{1-\gamma}}{1-\gamma}\right)$$

With $V_1 = \frac{nRT}{P}$ with $P = 101325 \ Pa$, $T = 300 \ K$, and $n = 1$ mole, $V = 0.0246 m^3$. This renders $V_2 = 0.0123 m^3$. $K = PV^{\gamma}, \therefore K = 210.83$.

After throwing that mess into a calculator, my final value is $1869.405\ J$ roughly.

Some questions of mine:

• Why can we not use $P = \frac{nRT}{V}$ as our function of $V$ for the integral instead of $P=\frac{K}{V^\gamma}$?

• My answer should not be negative, which should be due to $0.5 V_1 - V_1 = -0.5 V_1$, but I don't see how that logic was wrong. What's the problem there?

• Did I do anything laughably wrong here, and if so, where? I feel like I did something wrong.

• Shouldn't the 0.5V1 be in parenthesis? – Chet Miller Oct 26 '17 at 11:55

You can't directly integrate the ideal gas law because both the temperature and volume vary. You can solve for the temperature in terms of volume and then integrate it, but that just gives you the same $P=\frac{K}{V^\gamma}$.