A sample of gas with gamma = 1.5 is taken through an adiabatic process in which the volume is compre

Q: Why do we use $p_1 V_1^\gamma = p_2 V_2^\gamma$ instead of $p_1 V_1 = p_2 V_2$?

The equation $p_1 V_1 = p_2 V_2$ is Boyle's Law, which only applies to isothermal processes where temperature is constant. In an adiabatic process, temperature changes because no heat is allowed to enter or leave the system, so we must use Poisson's equation involving $\gamma$.

Q: How did $1200 \text{ cm}^3$ become $12 \times 10^{-4} \text{ m}^3$?

To convert from $\text{cm}^3$ to $\text{m}^3$, you must multiply by $(10^{-2})^3$, which is $10^{-6}$. So, $1200 \text{ cm}^3 = 1200 \times 10^{-6} \text{ m}^3$. Moving the decimal two places gives $12 \times 10^{-4} \text{ m}^3$.

Q: Why is the work done negative?

Work done by a gas is positive when it expands and negative when it is compressed. Since the volume decreased from $1200 \text{ cm}^3$ to $300 \text{ cm}^3$, the gas was compressed. This means work was done *on* the gas by an external force, resulting in a negative value.

Q: What does $\gamma = 1.5$ tell us about the gas?

The adiabatic index $\gamma$ is the ratio of specific heats ($C_p/C_v$). A value of $1.5$ doesn't perfectly match ideal monatomic ($1.67$) or diatomic ($1.4$) gases, but it represents a specific polyatomic gas mixture or a non-ideal state given for this problem's mathematical simplicity.

Q: Could we have used $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$?

Yes, absolutely! However, we weren't given the number of moles ($n$) or the temperatures ($T_1, T_2$). While we could use the ideal gas law to find them, using the pressure-volume form of the work equation is much faster and more direct.

Question

A sample of gas with $\gamma = 1.5$ is taken through an adiabatic process in which the volume is compressed from $1200 	ext{ cm}^3$ to $300 	ext{ cm}^3$. If the initial pressure is $200 	ext{ kPa}$. The absolute value of the work done by the gas in the process is ......... J.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a gas trapped inside a sturdy cylinder. We are about to compress it, and we are going to do it so rapidly that absolutely no heat has the time to escape into the surroundings. This is the hallmark of an **adiabatic process**.

In this specific scenario, our gas is being compressed from an initial volume of $1200 	ext{ cm}^3$ down to a final volume of $300 	ext{ cm}^3$. We are also given that the initial pressure is $200 	ext{ kPa}$, and the adiabatic index $\gamma$ is $1.5$.

Our ultimate goal is to find the absolute value of the work done during this intense compression. But before we can calculate the work, we are missing a crucial piece of the puzzle: the final pressure.

### The Master Equation: Poisson's Law

Because no heat is exchanged, Boyle's Law ($p_1 V_1 = p_2 V_2$) goes out the window. Instead, the pressure and volume are governed by **Poisson's Equation** for adiabatic processes:

$$p_1 V_1^\gamma = p_2 V_2^\gamma$$

This equation tells us that as the volume decreases, the pressure doesn't just increase linearly; it shoots up exponentially based on the power of $\gamma$. Let's rearrange this to isolate our unknown final pressure, $p_2$:

$$p_2 = p_1 \left(\frac{V_1}{V_2}ight)^\gamma$$

### Executing the Pressure Calculation

Now, let's substitute our known values into the rearranged equation. The initial pressure $p_1$ is $2 	imes 10^5 	ext{ Pa}$. The volume ratio is beautifully simple:

$$p_2 = 2 	imes 10^5 \left(\frac{1200}{300}ight)^{1.5}$$

Notice how we didn't need to convert the volumes to cubic meters just yet, because the units in the ratio perfectly cancel each other out! The fraction $\frac{1200}{300}$ simplifies to exactly $4$.

Now, we must evaluate $4^{1.5}$. A fractional power of $1.5$ is the same as $\frac{3}{2}$, which means we take the square root of $4$ (which is $2$) and then cube it ($2^3 = 8$).

$$p_2 = 2 	imes 10^5 	imes 8$$

$$p_2 = 16 	imes 10^5 	ext{ Pa}$$

The pressure has skyrocketed from $2 	imes 10^5 	ext{ Pa}$ to $16 	imes 10^5 	ext{ Pa}$. This massive spike is exactly what we expect when compressing a gas adiabatically!

### Calculating the Work Done

With both initial and final states fully known, we can now calculate the work done. The formula for work done in an adiabatic process represents the area under the $p-V$ curve:

$$W = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$$

Here is where we must be incredibly careful with our units. **Silly mistakes happen here!** We must convert our volumes from $	ext{cm}^3$ to standard SI units of $	ext{m}^3$ by multiplying by $10^{-6}$.

- $V_1 = 1200 	imes 10^{-6} = 12 	imes 10^{-4} 	ext{ m}^3$
- $V_2 = 300 	imes 10^{-6} = 3 	imes 10^{-4} 	ext{ m}^3$

Let's plug everything into our work formula:

$$W = \frac{(2 	imes 10^5)(12 	imes 10^{-4}) - (16 	imes 10^5)(3 	imes 10^{-4})}{1.5 - 1}$$

### The Final Calculation

Let's compute the numerator term by term. For the initial state, $(2 	imes 12) 	imes 10^{5-4} = 24 	imes 10^1 = 240$. For the final state, $(16 	imes 3) 	imes 10^{5-4} = 48 	imes 10^1 = 480$.

$$W = \frac{240 - 480}{0.5}$$

$$W = \frac{-240}{0.5}$$

Dividing by $0.5$ is mathematically identical to multiplying by $2$.

$$W = -480 	ext{ J}$$

The negative sign is physically significant. It tells us that the gas did not do work on its surroundings; rather, the surroundings did work **on the gas** to force it into a smaller volume.

However, the question specifically asks for the **absolute value** of the work done.

$$|W| = 480 	ext{ J}$$

And there we have it! A beautiful demonstration of adiabatic dynamics.

Enter Numerical Value:

Visualized Solution

Visualizing Adiabatic Compression

Poisson's Equation for Adiabatic Process

Substituting Values for $p_{2}$

Computing the Volume Ratio

Finalizing the Final Pressure $p_{2}$

Formula for Adiabatic Work Done

Substituting Values into Work Formula

Computing the Numerator

Finalizing the Work Done $W$

Absolute Value of Work Done

The Way Forward

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Executing the Pressure Calculation

Calculating the Work Done

The Final Calculation

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