During an experiment, an ideal gas is found to obey an additional law p^2V = constant. The gas is in

Q: Why do we eliminate pressure from the equation?

The question provides the initial and final states in terms of volume and asks for the final temperature. Therefore, we need a direct mathematical relationship between temperature ($T$) and volume ($V$). Eliminating pressure ($p$) bridges that gap.

Q: What happens to the $n$ and $R$ terms when we substitute?

Since the gas is enclosed, the number of moles ($n$) is constant. The universal gas constant ($R$) is also a constant. When we square them and move them to the other side of the equation, a constant divided by another constant is simply a new constant. We don't need their exact values to find the proportionality.

Q: Is this a polytropic process?

Yes! Any process that can be written in the form $pV^x = \text{constant}$ is a polytropic process. If we take the square root of the given law $p^2V = c$, we get $p V^{1/2} = c'$. So, this is a polytropic process with an index $x = 1/2$.

Q: Does the gas heat up or cool down during this expansion?

Since we found that $T \propto \sqrt{V}$, an increase in volume ($V$) will result in an increase in temperature ($T$). Therefore, the gas heats up as it expands in this specific process.

Q: Can we find the work done by the gas in this process?

Absolutely. Since it's a polytropic process with $x = 1/2$, you can use the standard formula for work done: $W = \frac{nR(T_{\text{final}} - T_{\text{initial}})}{1 - x}$. Substituting $x = 1/2$ and the temperatures will give you the exact work done.

Question

During an experiment, an ideal gas is found to obey an additional law $p^2V = 	ext{constant}$. The gas is initially at a temperature $T$ and volume $V$. When it expands to a volume $2V$, the temperature becomes ……… .

Sigma Percentile · Accepted Answer

### Unraveling a Custom Thermodynamic Process

Imagine you are conducting an experiment in a lab, and instead of following the usual isothermal or adiabatic paths, your gas decides to play by its own rules. In this problem, the gas obeys a unique law: **$p^2V = 	ext{constant}$**.

Our goal is to find out what happens to the temperature of this gas when it expands to twice its original volume. Let's break down the physics behind this expansion.

### The Master Equation

We are given a relationship between pressure ($p$) and volume ($V$), but we need to find how temperature ($T$) changes with volume. This is a classic scenario where the **Ideal Gas Equation** comes to the rescue.

For any ideal gas, we know that:
$$pV = nRT$$

To eliminate pressure from our custom law, we can rearrange the ideal gas equation to express pressure in terms of temperature and volume:
$$p = \frac{nRT}{V}$$

Now, let's substitute this expression for $p$ back into the given custom law:
$$\left(\frac{nRT}{V}ight)^2 V = 	ext{constant}$$

### Simplifying the Relation

Let's expand the squared term and see what we get:
$$\frac{n^2 R^2 T^2}{V^2} V = 	ext{constant}$$

Notice that $n$ (number of moles) and $R$ (universal gas constant) are constants for our enclosed gas. We can move them to the right side of the equation, where they simply merge into a new, generic constant. Also, one $V$ in the numerator cancels out with one $V$ in the denominator:
$$\frac{T^2}{V} = \frac{	ext{constant}}{n^2 R^2} = 	ext{new constant}$$

This leaves us with a beautiful, direct proportionality:
$$T^2 \propto V$$

Taking the square root of both sides, we find that the temperature is directly proportional to the square root of the volume:
$$T \propto \sqrt{V}$$

### The Final Calculation

Now that we have the relationship between temperature and volume, finding the final temperature is a breeze. We can set up a ratio for the initial and final states:
$$\frac{T_{	ext{final}}}{T_{	ext{initial}}} = \sqrt{\frac{V_{	ext{final}}}{V_{	ext{initial}}}}$$

We are given that the gas expands to twice its initial volume, so $V_{	ext{final}} = 2V$. Plugging this in:
$$\frac{T_{	ext{final}}}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$$

Multiplying both sides by $T$, we get our final answer:
$$T_{	ext{final}} = \sqrt{2}T$$

And there you have it! By combining the specific process law with the universal ideal gas equation, we successfully predicted the thermal behavior of the gas.

Animated Solution for Physics - Thermodynamics: During an experiment, an ideal gas is found to obey an additional law $p^{2} V = constant$ . The gas is initially at a temperature $T$ and volume $V$ . When it expands to a volume $2 V$ , the temperature becomes ……… .

Visualized Solution

Visualizing the Setup

The Custom Law

Ideal Gas Equation

Substituting Pressure

Simplifying the Relation

Proportionality

Final Calculation

The Way Forward

The Sigma Insight: Thermodynamic Processes

Unraveling a Custom Thermodynamic Process

The Master Equation

Simplifying the Relation

The Final Calculation

Similar Questions

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Visualized Solution

Visualizing the Setup

The Custom Law

Ideal Gas Equation

Substituting Pressure

Simplifying the Relation

Proportionality

Final Calculation

The Way Forward

The Sigma Insight: Thermodynamic Processes

Unraveling a Custom Thermodynamic Process

The Master Equation

Simplifying the Relation

The Final Calculation

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