A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is p.

Q: Why is the molar mass given if we don't use it?

It's a classic distractor! In competitive exams, extra information is often provided to test your conceptual clarity. Since the ideal gas law relates pressure to the number of moles ($n$), the specific identity or molar mass of the gas doesn't matter once $n$ is known.

Q: What if the question had given the mass of the gases instead of moles?

Great question! If the masses were given (e.g., $1 \text{ gram}$ of each gas), then you would absolutely need the molar mass to calculate the number of moles using $n = \frac{\text{mass}}{\text{molar mass}}$. In that case, the pressures would be different.

Q: Does the volume change when the temperature is doubled?

No, the problem states that the vessels are 'identical,' which implies they are rigid containers with the same fixed volume $V$.

Q: Is the ideal gas law applicable at such high temperatures?

For the scope of this problem, we assume ideal behavior. In reality, at extremely high temperatures or pressures, real gases deviate from ideal behavior, but here, the ideal gas model is perfectly sufficient.

Q: Why is pressure directly proportional to temperature?

According to the kinetic theory of gases, temperature is a measure of the average kinetic energy of the gas molecules. When temperature increases, molecules move faster and hit the container walls harder and more frequently, resulting in higher pressure.

Question

A vessel contains $1$ mole of $O_2$ gas (molar mass $32$) at a temperature $T$. The pressure of the gas is $p$. An identical vessel containing one mole of the gas (molar mass $4$) at a temperature $2T$ has a pressure of

Options:

(A) $\frac{p}{8}$

(B) $p$

(C) $2p$

(D) $8p$

Sigma Percentile · Accepted Answer

### The Setup: Two Vessels, Two Gases

Imagine you are in a laboratory, standing in front of two identical, sealed steel containers. The first container holds exactly one mole of oxygen gas at a temperature $T$, exerting a pressure $p$ on the walls.

The second container holds a mystery gas—likely helium, given its molar mass of 4—also exactly one mole, but heated to double the temperature, $2T$. We need to determine the pressure inside this second vessel.

### The Trap of Molar Mass

The question throws in the molar masses of the gases: 32 for oxygen and 4 for the mystery gas. **This is a classic misdirection!**

The ideal gas law, $pV = nRT$, relates pressure, volume, temperature, and the number of moles. It does not care about the specific identity or the molar mass of the gas, as long as we are dealing with the number of moles directly. Since we are already given that both vessels contain exactly 1 mole of gas, the molar mass is completely irrelevant to our calculation.

### The Master Equation

Let's bring in our trusty ideal gas equation:
$$p = \frac{nRT}{V}$$

Now, let's identify our constants. Both vessels have exactly one mole of gas, so $n$ is constant. They are identical vessels, meaning the volume $V$ is also constant. And $R$ is the universal gas constant.

Since $n$, $R$, and $V$ are all constants, we can establish a direct proportionality:
$$p \propto T$$

**The pressure of an ideal gas at constant volume and constant number of moles is directly proportional to its absolute temperature.**

### Final Calculation

In our second vessel, the temperature is $2T$, which is exactly double the initial temperature.

Because of the direct proportionality, if the temperature changes by a factor of 2, the pressure must also change by the exact same factor.

Therefore, the new pressure is simply double the initial pressure.
**The pressure in the second vessel is $2p$.**

This is a straightforward application of the ideal gas law, reminding us to focus only on the variables that actually matter and to avoid falling for unnecessary information!

Animated Solution for Physics - Thermodynamics: A vessel contains $1$ mole of $O_{2}$ gas (molar mass $32$ ) at a temperature $T$ . The pressure of the gas is $p$ . An identical vessel containing one mole of the gas (molar mass $4$ ) at a temperature $2 T$ has a pressure of

Select Answer:

Visualized Solution

$Analyzing the Setup$

$The Ideal Gas Equation$

$Identifying Constants$

$Proportionality$

$Final Calculation$

$Conclusion$

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Trap of Molar Mass

The Master Equation

Final Calculation

Similar Questions

The volume $V$ of an enclosure contains a mixture of three gases, $16 g$ of oxygen, $28 g$ of nitrogen and $44 g$ of carbon dioxide at absolute temperature $T$ . Consider $R$ as universal gas constant. The pressure of the mixture of gases is

A cylindrical container of volume $4.0 \times 1 0^{- 3} m^{3}$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is $400 K$ . The pressure of the mixture of gases is [Take, gas constant $= 8.3 J mol^{- 1} K^{- 1}$ ]

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2 : 3$ . The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is $4 : 3$ . The ratio of their densities is

For a given gas at $1 atm$ pressure, rms speed of the molecules is $200 m/s$ at $12 7^{\circ} C$ . At $2 atm$ pressure and at $22 7^{\circ} C$ , the rms speed of the molecules will be

A gas mixture consists of $2$ moles of oxygen and $4$ moles of argon at temperature $T$ . Neglecting all vibrational modes, the total internal energy of the system is

Consider a sample of oxygen behaving like an ideal gas. At $300 K$ , the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be (Molecular weight of oxygen is $32 g/mol$ ; $R = 8.3 J K^{- 1} mol^{- 1}$ )

Two moles of an ideal gas with $\frac{C _{p}}{C _{V}} = \frac{5}{3}$ are mixed with 3 mol of another ideal gas with $\frac{C _{p}}{C _{V}} = \frac{4}{3}$ . The value of $\frac{C _{p}}{C _{V}}$ for the mixture is

The temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ the root mean square velocity of the gas molecules is $v$ , at $480 K$ it becomes

Select Answer:

Visualized Solution

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

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

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

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

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

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