Two moles of an ideal monoatomic gas initially at pressure p1 and volume V1 undergo an adiabatic com

Q: Why is the work done in process B to C zero?

Work done by a gas is given by the area under the $p-V$ curve, or mathematically, $\int p \, dV$. Since the volume is constant during process $B \rightarrow C$ ($dV = 0$), the gas neither expands nor compresses, so it does zero work.

Q: How did we get $\gamma = 5/3$?

For an ideal monoatomic gas, the degrees of freedom $f = 3$ (only translational kinetic energy). The ratio of specific heats $\gamma$ is given by $1 + \frac{2}{f}$, which equals $1 + \frac{2}{3} = \frac{5}{3}$.

Q: Why is $\Delta U_{AB} = -W_{AB}$?

According to the First Law of Thermodynamics, $\Delta U = Q - W$. Since process $A \rightarrow B$ is adiabatic, there is no heat exchange ($Q = 0$). Therefore, the change in internal energy is exactly the negative of the work done by the gas.

Q: Why do we use $n C_V \Delta T$ for the total change in internal energy if the process isn't entirely at constant volume?

Internal energy $U$ is a state function, meaning its change depends ONLY on the initial and final temperatures, regardless of the path taken. For any ideal gas, $\Delta U$ is always equal to $n C_V \Delta T$ for any process.

Q: How did the $-\frac{3}{2} p_1 V_1$ term cancel out in the final temperature calculation?

When we expanded $n C_V (T_C - T_A)$, the $T_A$ term became $\frac{p_1 V_1}{2R}$. Multiplying this by $n C_V = 3R$ gave exactly $-\frac{3}{2} p_1 V_1$ on the left side. This perfectly matched the $-\frac{3}{2} p_1 V_1$ generated by the $-1$ in the work expression on the right side, allowing them to cancel out beautifully!

Question

Two moles of an ideal monoatomic gas initially at pressure $p_1$ and volume $V_1$ undergo an adiabatic compression until its volume is $V_2$. Then the gas is given heat $Q$ at constant volume $V_2$.
(a) Sketch the complete process on a $p-V$ diagram.
(b) Find the total work done by the gas, the total change in internal energy and the final temperature of the gas.
(Give your answer in terms of $p_1, V_1, V_2, Q$ and $R$)

Sigma Percentile · Accepted Answer

### The Thermodynamic Journey Begins

Imagine you are a scientist observing a sealed chamber containing exactly two moles of an ideal monoatomic gas. The gas is initially resting at a state we'll call point $A$, characterized by a pressure $p_1$ and a volume $V_1$.

Suddenly, the piston is pushed down rapidly, compressing the gas to a smaller volume $V_2$. Because this happens so fast, no heat can escape or enter the system. This is an **adiabatic compression**.

Once the gas is compressed, the piston is locked in place. The volume is now fixed at $V_2$. Then, a heat source is applied, pumping an amount of heat $Q$ into the gas. Since the volume cannot change, the pressure skyrockets. This is an **isochoric heating** process, taking us to our final state, point $C$.

### Mapping the Path: The $p-V$ Diagram

To truly understand what's happening, we must visualize it. The $p-V$ diagram is the map of our thermodynamic journey.

We start at point $A(V_1, p_1)$. The adiabatic compression traces a steep curve upwards and to the left, landing at point $B(V_2, p_B)$. Why is it steep? Because in an adiabatic compression, the pressure rises not just because the volume decreases, but also because the temperature increases!

From point $B$, the isochoric heating is represented by a perfectly vertical line shooting straight up to point $C(V_2, p_C)$. The volume is locked at $V_2$, but the added heat $Q$ causes the pressure to rise even further.

### Calculating the Work Done: The Adiabatic Squeeze

Now, let's calculate the total work done by the gas. Work is the area under the $p-V$ curve.

During the isochoric process ($B ightarrow C$), the volume doesn't change ($dV = 0$). Therefore, the gas does absolutely zero work:
$$W_{BC} = 0$$

All the work is done during the adiabatic compression ($A ightarrow B$). The formula for work done in an adiabatic process is:
$$W_{AB} = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$$

But there's a catch! We don't know the intermediate pressure $p_2$ at point $B$. We must find it using the adiabatic condition $p_1 V_1^\gamma = p_2 V_2^\gamma$, which gives us:
$$p_2 = p_1 \left(\frac{V_1}{V_2}ight)^\gamma$$

Substituting this back into our work equation, we can factor out $p_1 V_1$:
$$W_{AB} = \frac{p_1 V_1 \left[ 1 - \left(\frac{V_1}{V_2}ight)^{\gamma-1} ight]}{\gamma - 1}$$

For a monoatomic gas, the ratio of specific heats $\gamma$ is $\frac{5}{3}$. This means $\gamma - 1 = \frac{2}{3}$. Plugging this in, we get the total work done:
$$W_{	ext{Total}} = -\frac{3}{2} p_1 V_1 \left[ \left(\frac{V_1}{V_2}ight)^{2/3} - 1 ight]$$

Notice the negative sign? It perfectly reflects the physical reality: the gas was compressed, meaning work was done **on** the gas, not **by** it.

### Unlocking the Internal Energy

Next, we seek the total change in internal energy, $\Delta U_{	ext{Total}}$. The First Law of Thermodynamics, $\Delta U = Q - W$, is our guiding light here.

Let's break it down step-by-step. During the adiabatic compression ($A ightarrow B$), no heat is exchanged ($Q_{AB} = 0$). Thus, the change in internal energy is simply the negative of the work done:
$$\Delta U_{AB} = -W_{AB}$$

During the isochoric heating ($B ightarrow C$), no work is done ($W_{BC} = 0$). The internal energy increases by exactly the amount of heat added:
$$\Delta U_{BC} = Q$$

The total change in internal energy is the sum of these two parts:
$$\Delta U_{	ext{Total}} = \Delta U_{AB} + \Delta U_{BC} = -W_{AB} + Q$$

Substituting our work expression, the negative sign flips, giving us:
$$\Delta U_{	ext{Total}} = \frac{3}{2} p_1 V_1 \left[ \left(\frac{V_1}{V_2}ight)^{2/3} - 1 ight] + Q$$

### The Final Destination: Finding the Temperature

Finally, we want to find the temperature of the gas at the end of its journey, $T_C$.

Internal energy is a state function. Its total change depends only on the initial and final temperatures, regardless of the crazy path we took to get there! For any ideal gas, this change is given by:
$$\Delta U_{	ext{Total}} = n C_V (T_C - T_A)$$

We have $n = 2$ moles. For a monoatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$. Therefore, $n C_V = 3R$.

What about the initial temperature $T_A$? We can easily find it using the ideal gas law at point $A$:
$$T_A = \frac{p_1 V_1}{nR} = \frac{p_1 V_1}{2R}$$

Now, we set up our master equation:
$$3R \left( T_C - \frac{p_1 V_1}{2R} ight) = \frac{3}{2} p_1 V_1 \left[ \left(\frac{V_1}{V_2}ight)^{2/3} - 1 ight] + Q$$

Let's expand the left side:
$$3R T_C - \frac{3}{2} p_1 V_1 = \frac{3}{2} p_1 V_1 \left(\frac{V_1}{V_2}ight)^{2/3} - \frac{3}{2} p_1 V_1 + Q$$

Look at the magic of algebra! The $-\frac{3}{2} p_1 V_1$ term appears on both sides and cancels out beautifully. We are left with:
$$3R T_C = \frac{3}{2} p_1 V_1 \left(\frac{V_1}{V_2}ight)^{2/3} + Q$$

Dividing everything by $3R$, we arrive at our final, elegant answer:
$$T_C = \frac{Q}{3R} + \frac{p_1 V_1}{2R} \left(\frac{V_1}{V_2}ight)^{2/3}$$

And there you have it! By carefully tracking the work and heat through each step of the process, we've completely solved the thermodynamic state of the gas.

Visualized Solution

\text{Initial Setup \& Process Identification}

\text{The } p-V \text{ Diagram}

\text{Work Done in Process } A \rightarrow B

\text{Calculating Total Work Done}

\text{Change in Internal Energy}

\text{Total Internal Energy Expression}

\text{Final Temperature } (T_C)

\text{Solving for } T_C

The Sigma Insight: Thermodynamic Processes

The Thermodynamic Journey Begins

Mapping the Path

Calculating the Work Done

Unlocking the Internal Energy

The Final Destination

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