A gas has a compressibility factor of 0.5 and a molar volume of 0.4 dm3 mol-1 at a temperature of 80

Q: Why did we take $1 \text{ dm}^3$ as $1 \text{ L}$?

By definition, 1 liter is the volume of a cube that is 10 cm on each side. Since 10 cm is 1 decimeter (dm), the volume is $1 \text{ dm} \times 1 \text{ dm} \times 1 \text{ dm} = 1 \text{ dm}^3$. Therefore, $1 \text{ L} = 1 \text{ dm}^3$.

Q: Can we use $R = 0.0821$ instead of $0.08$?

The problem explicitly provides the value of the gas constant as $R = 8 \times 10^{-2} \text{ L atm K}^{-1} \text{ mol}^{-1}$, which is exactly $0.08$. Using $0.0821$ will lead to a slightly different answer and might cause you to lose marks in an integer type question. Always use the given constants!

Q: What does a compressibility factor $Z = 0.5$ physically mean?

The compressibility factor $Z$ measures how much a real gas deviates from ideal behavior. When $Z < 1$, it means the gas is more compressible than an ideal gas. This happens because attractive intermolecular forces are dominating, pulling the molecules closer together and reducing the volume.

Q: Why is $Z = \frac{V_{\text{real}}}{V_{\text{ideal}}}$?

By definition, $Z = \frac{P V_{\text{real}}}{RT}$. For an ideal gas at the same pressure and temperature, $V_{\text{ideal}} = \frac{RT}{P}$. Substituting $\frac{RT}{P}$ with $V_{\text{ideal}}$ in the first equation gives $Z = \frac{V_{\text{real}}}{V_{\text{ideal}}}$. It's a very handy shortcut!

Q: What if the question asked for the ratio $y/x$?

You would simply invert the final result. Since $x/y = 100$, the ratio $y/x$ would be $1/100 = 0.01$. Always read carefully to see which ratio is being asked!

Question

A gas has a compressibility factor of 0.5 and a molar volume of $0.4 	ext{ dm}^3 	ext{ mol}^{-1}$ at a temperature of 800 K and pressure x atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be y $	ext{dm}^3 	ext{ mol}^{-1}$. The value of x/y is ______.

[Use: Gas constant, $	ext{R} = 8 	imes 10^{-2} 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a mysterious gas trapped in a container. We are given a fascinating scenario where we observe this gas in two different theoretical states. First, it behaves as a **real gas** with a compressibility factor ($Z$) of $0.5$. Then, we are asked to imagine it behaving **ideally** at the exact same temperature and pressure.

Before we dive into the equations, let's clear up a small unit detail. The volume is given in $	ext{dm}^3 	ext{ mol}^{-1}$. It is crucial to remember that $1 	ext{ dm}^3$ is exactly equal to $1 	ext{ L}$. Therefore, our real molar volume is simply $0.4 	ext{ L mol}^{-1}$.

### The Master Equation

To find the unknown pressure $x$, we need our master tool: the **compressibility factor equation**. This formula acts as the bridge between real and ideal behavior.

$$Z = \frac{P V_m}{R T}$$

Let's carefully substitute our known values into this equation. We know $Z = 0.5$, the molar volume $V_m = 0.4 	ext{ L mol}^{-1}$, and the temperature $T = 800 	ext{ K}$. For the gas constant $R$, the problem explicitly tells us to use $8 	imes 10^{-2}$, which is $0.08 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$.

$$0.5 = \frac{x 	imes 0.4}{0.08 	imes 800}$$

Now, let's do the math. Multiplying the denominator, $0.08 	imes 800$ gives us $64$.

$$0.5 = \frac{0.4 x}{64}$$

Multiplying $64$ by $0.5$ gives $32$. Dividing $32$ by $0.4$, we get our pressure $x$:

$$x = 80 	ext{ atm}$$

### The Ideal Shortcut

Next, we need to find the ideal molar volume, $y$. For an ideal gas, the volume is simply $R T$ divided by $P$. We could plug our newly found pressure of $80 	ext{ atm}$ back into the ideal gas equation.

$$y = \frac{0.08 	imes 800}{80} = \frac{64}{80} = 0.8 	ext{ L mol}^{-1}$$

But wait, there's a beautiful shortcut! The compressibility factor $Z$ is also exactly equal to the ratio of real volume to ideal volume at the same temperature and pressure.

$$Z = \frac{V_{	ext{real}}}{V_{	ext{ideal}}}$$

Using our shortcut, $0.5 = \frac{0.4}{y}$. Solving this, we instantly get $y = 0.8 	ext{ L mol}^{-1}$. Both methods yield the exact same result, but the shortcut is a powerful tool to save time during an exam!

### Final Calculation

We are almost at the finish line. The question asks for the ratio of $x$ to $y$. We simply divide our pressure, $80$, by our ideal volume, $0.8$.

$$\frac{x}{y} = \frac{80}{0.8} = 100$$

And there we have it! Our final answer is **$100$**. As a quick conceptual check, since $Z < 1$, it means the real gas is more compressible than an ideal gas, indicating that attractive intermolecular forces are dominating in this state.

Enter Numerical Value:

Visualized Solution

State Variables of the Gas

Compressibility Factor $Z$

Substituting to find $x$

Calculating Pressure $x$

Ideal Gas Molar Volume $y$

Calculating Volume $y$

Final Ratio $x / y$

Conclusion

The Sigma Insight: Gaseous State

Analyzing the Setup

The Master Equation

The Ideal Shortcut

Final Calculation

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Enter Numerical Value:

Visualized Solution

State Variables of the Gas

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

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

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

Ideal Gas Molar Volume .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

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

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