The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than

Q: What exactly is the compressibility factor?

The compressibility factor, denoted by $Z$, is a correction factor that describes how much a real gas deviates from ideal gas behavior. It is defined as the ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure, which simplifies to $Z = \frac{pV}{nRT}$.

Q: Why did we assume $R$ is the same for both gases?

$R$ is the universal gas constant. As the name implies, it is a fundamental physical constant that has the exact same value for all gases, regardless of their chemical nature.

Q: Can we just assume the gases are ideal to solve this faster?

No, we cannot. If the gases were ideal, their compressibility factors would both be exactly $1$ ($Z_A = 1$ and $Z_B = 1$). This would contradict the given condition that $Z_A = 3Z_B$. The problem inherently deals with real gases.

Q: What if the number of moles were not equal?

If the moles were different, you would simply keep $n_A$ and $n_B$ in the equation. The final relationship between the pressures would then depend on the ratio of their moles, resulting in an equation like $2p_A n_B = 3p_B n_A$.

Q: Why did we substitute $V_A$ with $2V_B$?

The problem explicitly states that 'The volume of gas A is twice than that of gas B'. Mathematically, this translates directly to the equation $V_A = 2 \times V_B$.

Question

The volume of gas $A$ is twice than that of gas $B$. The compressibility factor of gas $A$ is thrice than that of gas $B$ at same temperature. The pressures of the gases for equal number of moles are

Options:

(A) $p_A = 2p_B$

(B) $2p_A = 3p_B$

(C) $p_A = 3p_B$

(D) $3p_A = 2p_B$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine we have two distinct containers, one holding Gas A and the other holding Gas B. The problem provides us with a very specific set of constraints. First, the volume of Gas A is exactly twice the volume of Gas B, which we can write mathematically as **$V_A = 2V_B$**.

Furthermore, we are told that both containers hold an equal number of moles of gas ($n_A = n_B$) and are maintained at the exact same temperature ($T_A = T_B$). This sets the stage for a direct comparison between the two gases.

### The Master Equation
The core of this problem revolves around the **compressibility factor**, denoted by $Z$. Do you remember the formula for the compressibility factor? It is a thermodynamic property that describes the deviation of a real gas from ideal gas behavior. Mathematically, it is defined as:

$$Z = \frac{pV}{nRT}$$

The problem explicitly states a relationship between the compressibility factors of the two gases: the compressibility factor of Gas A is three times that of Gas B.

$$Z_A = 3Z_B$$

### Substitution and Simplification
Now, let's substitute our formula for $Z$ into this relationship for both gases. This gives us:

$$\frac{p_A V_A}{n_A R T_A} = 3 \frac{p_B V_B}{n_B R T_B}$$

Next, we carefully substitute the constraints we identified earlier. We replace $V_A$ with $2V_B$. Since the moles and temperatures are equal, we can simply use $n_B$ and $T_B$ on the left side of the equation as well.

$$\frac{p_A (2V_B)}{n_B R T_B} = 3 \frac{p_B V_B}{n_B R T_B}$$

Look closely at this equation. It's time for the satisfying part—canceling out the common terms! The volume $V_B$, the number of moles $n_B$, the universal gas constant $R$, and the temperature $T_B$ are present in the denominators and numerators on both sides.

### Final Calculation
After slashing out all these common terms, the clutter clears, and we are left with a beautifully simple and elegant relationship:

**$$2p_A = 3p_B$$**

This tells us exactly how the pressures of the two gases relate to each other under these specific conditions. It's a great reminder of how powerful systematic substitution and algebraic simplification can be in physical chemistry!

Animated Solution for Chemistry - States of Matter: The volume of gas $A$ is twice than that of gas $B$ . The compressibility factor of gas $A$ is thrice than that of gas $B$ at same temperature. The pressures of the gases for equal number of moles are

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Visualized Solution

$V_{A} = 2 V_{B}, n_{A} = n_{B}, T_{A} = T_{B}$

$Z = \frac{p V}{n R T}$

$Z_{A} = 3 Z_{B}$

$Substitution$

$Simplification$

$2 p_{A} = 3 p_{B}$

$Ideal Gas Scenario$

The Sigma Insight: Gaseous State

Analyzing the Setup Imagine we have two distinct containers, one holding Gas A and the other holding Gas B

The Master Equation The core of this problem revolves around the compressibility factor, denoted by $Z$

Substitution and Simplification Now, let's substitute our formula for $Z$ into this relationship for both gases

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ZA​=3ZB​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2pA​=3pB​

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