Animated Solution for Physics - Thermodynamics: Two moles of an ideal monoatomic gas occupies a volume $V$ at $27^{\circ }\text{C}$. The gas expands adiabatically to a volume $2V$. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.

A Gas in Isolation Imagine a perfectly insulated cylinder containing $2$ moles of an ideal monoatomic gas. The gas is initially at a comfortable room temperature of $27^{\circ }\text{C}$, which translates to $300\text{ K}$ on the absolute scale. It occupies a volume $V$. Suddenly, the piston is released, and the gas expands to double its volume, reaching $2V$. Because the cylinder is insulated, no heat can enter or leave the system. This is the hallmark of an adiabatic process.

Temperature and Volume In an adiabatic process, the pressure, volume, and temperature are intricately linked. While Boyle's Law ($PV=\text{constant}$) works for isothermal processes, adiabatic expansion follows a different rule because the temperature drops as the gas does work. The relationship between temperature and volume is given by Poisson's equation:

Here, $\gamma$ is the adiabatic index, which is the ratio of specific heats ($C_p/C_v$). For a monoatomic gas like Helium or Argon, $\gamma =\frac{5}{3}$.

Crunching the Numbers Let's substitute our known values into the master equation. We know $T_1=300\text{ K}$, $V_1=V$, and $V_2=2V$. The exponent becomes $\frac{5}{3}-1=\frac{2}{3}$.

Notice how beautifully the $V^{\frac{2}{3}}$ terms cancel out on both sides. We are left with a pure numerical calculation:

Calculating $2^{\frac{2}{3}}$ (which is the cube root of $4$) gives approximately $1.587$. Dividing $300$ by this value yields our final temperature:

The gas has cooled down significantly! This happens because the gas expended its own internal energy to push the piston outward.

Paying for Expansion Now, let's quantify exactly how much energy the gas spent. The change in internal energy ($\Delta U$) for any ideal gas process depends solely on the change in temperature:

For a monoatomic gas, the molar heat capacity at constant volume is $C_v=\frac{3}{2}R$. Let's plug in our values ($n=2$, $\Delta T=189-300=-111\text{ K}$):

Converting this to kilojoules, we get $\Delta U\approx -2.7\text{ kJ}$.

It tells us that the system lost energy. Since no heat was added ($\Delta Q=0$), the First Law of Thermodynamics ($\Delta Q=\Delta U+\Delta W$) dictates that $\Delta U=-\Delta W$. The gas did $+2.7\text{ kJ}$ of work on the surroundings, paying for it entirely from its internal thermal reservoir. Thus, the final temperature is $189\text{ K}$, and the change in internal energy is $-2.7\text{ kJ}$.