One mole of an ideal gas passes through a process, where pressure and volume obey the relation p = p

Q: Why can't we just use $\Delta T = \frac{p \Delta V}{R}$?

Because pressure is not constant! The pressure changes as the volume changes according to the given relation. You must find the exact temperature at each state before subtracting.

Q: What happened to the number of moles $n$ in the ideal gas equation?

The problem explicitly states 'One mole of an ideal gas', so we substitute $n = 1$ into $pV = nRT$, giving us $pV = RT$.

Q: How did $\frac{1}{2} \left( \frac{V_0}{2V_0} \right)^2$ become $\frac{1}{8}$?

First, the $V_0$ terms cancel out inside the parenthesis, leaving $(1/2)^2$, which evaluates to $1/4$. Then, multiplying by the $1/2$ outside the parenthesis gives $1/8$.

Q: Is this an isothermal process?

No. In an isothermal process, temperature remains constant, so $\Delta T$ would be zero. Here, we clearly calculated a non-zero change in temperature, meaning the temperature varies throughout the process.

Q: Could we have found the change in temperature by integrating?

Integration is used to find the work done ($\int p \, dV$). To find the change in state variables like temperature, you only need the initial and final states, not the path integral.

Question

One mole of an ideal gas passes through a process, where pressure and volume obey the relation $p = p_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$. Here, $p_0$ and $V_0$ are constants. Calculate the change in the temperature of the gas, if its volume changes from $V_0$ to $2V_0$.

Options:

(A) $\frac{1}{2}\frac{p_0 V_0}{R}$

(B) $\frac{1}{4}\frac{p_0 V_0}{R}$

(C) $\frac{3}{4}\frac{p_0 V_0}{R}$

(D) $\frac{5}{4}\frac{p_0 V_0}{R}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracking the state of an ideal gas as it undergoes a specific thermodynamic process. The problem provides us with a unique, slightly intimidating relation between pressure and volume:

$$p = p_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$$

Our ultimate goal is to find the change in temperature as the gas expands from an initial volume of $V_0$ to a final volume of $2V_0$. But right now, we only have a relation between pressure and volume. How do we bring temperature into the picture?

### The Master Equation

The ideal gas equation is our universal translator here. It perfectly relates pressure, volume, and temperature. For one mole of gas ($n=1$), the equation is:

$$pV = RT \implies p = \frac{RT}{V}$$

Let's simply take the given expression for pressure and equate it with the pressure we just derived from the ideal gas equation:

$$\frac{RT}{V} = p_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$$

Now, we need to isolate our target variable, temperature $T$. By multiplying both sides of our equation by the volume $V$, we create a powerful new equation. This gives us a direct formula to calculate the exact temperature of the gas at any given volume during this entire process:

$$T(V) = \frac{p_0 V}{R} \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$$

### Calculating the Temperatures

Let's put our new formula to work. First, we need the initial temperature, let's call it $T_1$. This happens when the volume is exactly $V_0$. We carefully substitute $V_0$ into our temperature function:

$$T_1 = \frac{p_0 V_0}{R} \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V_0} \right)^2 \right]$$

Watch how beautifully the terms cancel out to give us a simple expression:

$$T_1 = \frac{p_0 V_0}{R} \left[ 1 - \frac{1}{2} \right] = \frac{1}{2} \frac{p_0 V_0}{R}$$

Next up is the final state. We need to calculate the final temperature, $T_2$, when the gas has expanded and the volume has doubled to $2V_0$. We must be very careful here to substitute $2V_0$ in place of $V$ everywhere in the equation, especially inside that squared term:

$$T_2 = \frac{p_0 (2V_0)}{R} \left[ 1 - \frac{1}{2} \left( \frac{V_0}{2V_0} \right)^2 \right]$$

Simplifying the fraction inside the square:

$$T_2 = \frac{2p_0 V_0}{R} \left[ 1 - \frac{1}{2} \left( \frac{1}{4} \right) \right]$$

$$T_2 = \frac{2p_0 V_0}{R} \left[ 1 - \frac{1}{8} \right] = \frac{2p_0 V_0}{R} \left[ \frac{7}{8} \right] = \frac{7}{4} \frac{p_0 V_0}{R}$$

### Final Calculation

We are in the endgame now. The change in temperature is simply the final temperature minus the initial temperature. We take our expression for $T_2$ and subtract $T_1$:

$$\Delta T = T_2 - T_1$$

$$\Delta T = \frac{7}{4} \frac{p_0 V_0}{R} - \frac{1}{2} \frac{p_0 V_0}{R}$$

A little bit of fraction math, and we arrive at our final, elegant answer:

$$\Delta T = \left( \frac{7}{4} - \frac{2}{4} \right) \frac{p_0 V_0}{R} = \mathbf{\frac{5}{4} \frac{p_0 V_0}{R}}$$

Select Answer:

Visualized Solution

Visualizing the Process

The Ideal Gas Equation

Equating Pressures

Temperature as a Function of Volume

Calculating Initial Temperature

Calculating Final Temperature

Change in Temperature

Food for Thought

The Sigma Insight: Kinetic Theory of Gases and Gas Laws

Analyzing the Setup

The Master Equation

Calculating the Temperatures

Final Calculation

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