At 27°C two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a v

Q: Why do we use $C_V$ for calculating $\Delta U$ even though the volume is changing?

Internal energy of an ideal gas depends *only* on its temperature. The formula $\Delta U = nC_V\Delta T$ is universally valid for any process of an ideal gas, not just isochoric ones.

Q: Why did the temperature of the gas drop during expansion?

In an adiabatic expansion, the gas does work on its surroundings without any heat entering the system ($Q=0$). It must use its own internal energy to do this work, which leads to a decrease in temperature.

Q: How do we know $\gamma = 5/3$?

The problem states it is a monoatomic gas. For monoatomic gases (like Helium or Argon), the degrees of freedom $f = 3$. The ratio of specific heats $\gamma = 1 + 2/f$, which gives $1 + 2/3 = 5/3$.

Q: Can we use $pV^\gamma = \text{constant}$ instead of $TV^{\gamma-1} = \text{constant}$?

Yes, you can! But you would first need to find the initial pressure using the ideal gas law, then find the final pressure, and finally use the ideal gas law again to find the final temperature. Using the $T-V$ relation is much faster here.

Q: Why is the work done positive while the change in internal energy is negative?

The gas is expanding, meaning its volume is increasing. By convention, work done *by* the gas during expansion is positive. Since $Q = 0$, the First Law ($Q = W + \Delta U$) dictates that $W = -\Delta U$. A positive $W$ perfectly balances a negative $\Delta U$.

Question

At $27^{\circ}	ext{C}$ two moles of an ideal monoatomic gas occupy a volume $V$. The gas expands adiabatically to a volume $2V$.
Calculate
(a) the final temperature of the gas,
(b) change in its internal energy,
(c) the work done by the gas during this process.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a gas trapped inside a cylinder fitted with a piston. We are given $2 	ext{ moles}$ of an ideal monoatomic gas initially at a temperature of $27^\circ	ext{C}$.

Before we dive into the math, let's convert this temperature into the absolute Kelvin scale.

$$T_1 = 27 + 273 = 300	ext{ K}$$

The gas is allowed to expand adiabatically until its volume doubles, meaning $V_2 = 2V$. The word **adiabatically** is the most crucial hint here. It tells us that the cylinder is perfectly insulated, and there is absolutely zero heat exchange with the surroundings ($Q = 0$).

Because it's a monoatomic gas, we also know its degrees of freedom $f = 3$, which gives us the ratio of specific heats $\gamma = \frac{5}{3}$.

### The Master Equation

For an adiabatic process, the pressure, volume, and temperature are constantly changing, but they follow specific conserved relationships. Since we are dealing with temperature and volume, we will use the $T-V$ relation:

$$TV^{\gamma-1} = 	ext{constant}$$

Let's set up the equation for our initial and final states:

$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$$

Now, we carefully substitute our known values into this master equation:

$$300 	imes V^{\frac{5}{3}-1} = T_2 	imes (2V)^{\frac{5}{3}-1}$$

### Final Calculation

Let's simplify the exponents. $\frac{5}{3} - 1 = \frac{2}{3}$.

$$300 	imes V^{\frac{2}{3}} = T_2 	imes (2V)^{\frac{2}{3}}$$

Notice how the $V^{\frac{2}{3}}$ term beautifully cancels out from both sides!

$$T_2 = 300 	imes \left(\frac{1}{2}ight)^{\frac{2}{3}}$$

Calculating this value gives us the final temperature:

$$T_2 \approx 189	ext{ K}$$

The gas has cooled down significantly! This makes perfect physical sense. Since the gas expanded and did work without any heat entering the system, it had to spend its own internal energy, causing its temperature to drop.

### Change in Internal Energy

Next, we need to find the change in internal energy ($\Delta U$). The internal energy of an ideal gas depends solely on its temperature. The universal formula for any process is:

$$\Delta U = nC_V\Delta T$$

For a monoatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$. Let's plug in our values:

$$\Delta U = 2 	imes \left(\frac{3}{2} 	imes 8.31ight) 	imes (189 - 300)$$

$$\Delta U = 3 	imes 8.31 	imes (-111)$$

$$\Delta U \approx -2767	ext{ J}$$

The negative sign confirms that the internal energy has decreased.

### The First Law of Thermodynamics

Finally, we need to calculate the work done by the gas. We invoke the First Law of Thermodynamics:

$$Q = W + \Delta U$$

Since the process is adiabatic, $Q = 0$.

$$0 = W + \Delta U$$

$$W = -\Delta U$$

$$W = -(-2767	ext{ J}) = 2767	ext{ J}$$

The work done is positive, which aligns with the fact that the gas expanded (volume increased). The gas performed $2767	ext{ J}$ of work on the surroundings entirely at the expense of its own internal energy!

Visualized Solution

\text{Initial Setup}

\text{Adiabatic Process Equation}

\text{Substituting Values}

\text{Calculating Final Temperature}

\text{Change in Internal Energy Formula}

\text{Calculating } \Delta U

\text{First Law of Thermodynamics}

\text{Calculating Work Done}

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Final Calculation

Change in Internal Energy

The First Law of Thermodynamics

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