Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pres

Q: Why is the work done negative?

In thermodynamics, work done *by* the gas is considered positive (expansion), and work done *on* the gas is negative (compression). Since the volume decreases from 6 L to 2 L, the gas is compressed, hence the negative work.

Q: How did we know to use $C_p - C_V = R$?

This is Mayer's relation, which holds true for all ideal gases. Since we were given $C_V$ and needed $\gamma = C_p / C_V$, finding $C_p$ was the logical intermediate step.

Q: Why do we need to convert liters to cubic meters?

The pressure is given in SI units ($\text{N/m}^2$ or Pascals). To ensure our final energy (work) is in the standard SI unit of Joules, all other quantities must also be in SI units. $1 \text{ L} = 10^{-3} \text{ m}^3$.

Q: Can we use $W = \frac{nR(T_i - T_f)}{\gamma - 1}$ instead?

Absolutely! But to do that, you would first need to calculate the initial and final temperatures using the ideal gas law ($pV = nRT$). Using the pressure-volume formula directly saves us that extra step.

Q: How to calculate $(3)^{5/3}$ without a calculator?

You can write it as $3^{1 + 2/3} = 3^1 \times 3^{2/3} = 3 \times (3^2)^{1/3} = 3 \times \sqrt[3]{9}$. Since $2^3 = 8$, $\sqrt[3]{9}$ is slightly more than 2 (around 2.08). So, $3 \times 2.08 \approx 6.24$.

Question

Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are $10^5 	ext{ N/m}^2$ and $6 	ext{ L}$ respectively. The final volume of the gas is $2 	ext{ L}$ molar specific heat of the gas at constant volume is $3R/2$.

Sigma Percentile · Accepted Answer

The problem of adiabatic compression is a classic in thermodynamics, testing our understanding of how pressure, volume, and temperature interact when no heat is exchanged with the surroundings. Let's dive into this elegant problem and unravel the physics step by step.

### Decoding the Adiabatic Compression

Imagine a gas trapped in a perfectly insulated cylinder. When we compress it, we are doing work on the gas. Since no heat can escape, all that work goes directly into increasing the internal energy of the gas, causing its temperature and pressure to spike dramatically.

We are given that the gas is compressed from an initial volume of $6 	ext{ L}$ to a final volume of $2 	ext{ L}$. The initial pressure is $10^5 	ext{ N/m}^2$. To find the work done, we first need to know the final pressure, and for that, we need the adiabatic index, $\gamma$.

### Unveiling the Adiabatic Index

The problem provides the molar specific heat at constant volume, $C_V = \frac{3R}{2}$. This immediately tells us we are dealing with a monoatomic gas.

To find the specific heat at constant pressure, $C_p$, we use Mayer's relation:

$$C_p = C_V + R = \frac{3R}{2} + R = \frac{5R}{2}$$

The adiabatic index $\gamma$ is simply the ratio of these two specific heats:

$$\gamma = \frac{C_p}{C_V} = \frac{5R/2}{3R/2} = \frac{5}{3}$$

### The Journey to Final Pressure

In an adiabatic process, the relationship between pressure and volume is governed by the equation $pV^\gamma = 	ext{constant}$. We can set up an equation relating the initial and final states:

$$p_i V_i^\gamma = p_f V_f^\gamma$$

Rearranging this to solve for the final pressure $p_f$, we get:

$$p_f = p_i \left(\frac{V_i}{V_f}ight)^\gamma$$

Now, we substitute the known values. The volume ratio $V_i / V_f$ is $6 / 2 = 3$.

$$p_f = 10^5 	imes (3)^{5/3}$$

Calculating $3^{5/3}$ might seem tricky, but it's just $3^1 	imes 3^{2/3} = 3 	imes \sqrt[3]{9} \approx 6.24$.

$$p_f \approx 6.24 	imes 10^5 	ext{ N/m}^2$$

### The Grand Finale: Calculating Work Done

With both the initial and final states fully defined, we can now calculate the work done. The formula for work done during an adiabatic process is derived from integrating $p \, dV$:

$$W = \frac{p_i V_i - p_f V_f}{\gamma - 1}$$

Before plugging in the numbers, we must ensure all units are consistent. The volumes must be converted from liters to cubic meters by multiplying by $10^{-3}$. Let's calculate the $pV$ products:

$$p_i V_i = 10^5 	imes 6 	imes 10^{-3} = 600 	ext{ J}$$

$$p_f V_f = 6.24 	imes 10^5 	imes 2 	imes 10^{-3} = 1248 	ext{ J}$$

The denominator is simply $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$. Now, substitute everything into the work formula:

$$W = \frac{600 - 1248}{2/3}$$

$$W = \frac{-648}{2/3} = -648 	imes \frac{3}{2}$$

$$W = -972 	ext{ J}$$

The final answer is **$-972 	ext{ J}$**. The negative sign is a beautiful confirmation of the physics: because the gas was compressed, work was done *on* the system, not *by* it.

Enter Numerical Value:

Visualized Solution

$G i v e n P a r am e t er s$

$F in d in g γ$

$A d iaba t i c P r ocess E q u a t i o n$

$C a l c u l a t in g F ina l P r ess u r e$

$W or k D o n e F or m u l a$

$S u b s t i t u t in g V a l u es$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Thermodynamic Processes

Decoding the Adiabatic Compression

Unveiling the Adiabatic Index

The Journey to Final Pressure

The Grand Finale

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding γ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Process Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Final Pressure

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation