NEET 2025 Physics Ideal Gas Equation MCQ Question

Let's break down the problem step-by-step and derive the mass of oxygen withdrawn from the cylinder using the ideal gas law.

Given Data:

• Volume of the cylinder, $V$ = 30 L = 30 × 10⁻³ m³ = 0.03 m³
• Initial moles of oxygen, $n_i$ = 18.20 moles
• Gauge pressure after withdrawal = 11 atm
• Temperature, $T = 27°C = 300 K$
• Gas constant, $R = \frac{100}{12} \, \text{J mol}^{-1} \text{K}^{-1} \approx 8.333 \, \text{J mol}^{-1} \text{K}^{-1}$
• 1 atm = $1.01 \times 10^5 \, \text{N/m}^2$

Step 1: Calculate the Final Absolute Pressure

Gauge pressure is defined as the pressure relative to atmospheric pressure. To find the absolute pressure, we add the atmospheric pressure to the gauge pressure:

$$P_{\text{final}} = P_{\text{gauge}} + P_{\text{atm}} = 11 \, \text{atm} + 1 \, \text{atm} = 12 \, \text{atm}$$

Now, convert this pressure to pascals:

$$P_{\text{final}} = 12 \, \text{atm} \times 1.01 \times 10^5 \, \text{N/m}^2 \approx 12.12 \times 10^5 \, \text{N/m}^2$$

Step 2: Use the Ideal Gas Law

According to the ideal gas law,

$$PV = nRT$$

We can rearrange this to solve for the number of moles after the withdrawal:

$$n_{\text{final}} = \frac{PV}{RT}$$

Substituting the values

Substituting the values we have:

$$n_{\text{final}} = \frac{(12.12 \times 10^5) \times 0.03}{\left(\frac{100}{12}\right) \times 300}$$

Calculating the numerator:

$$12.12 \times 10^5 \times 0.03 = 12.12 \times 3000 = 36360$$

Calculating the denominator:

$$\left(\frac{100}{12}\right) \times 300 = \frac{30000}{12} = 2500$$

Now, substituting back into the equation for $n_{\text{final}}$:

$$n_{\text{final}} = \frac{36360}{2500} \approx 14.544 \, \text{moles}$$

Step 3: Calculate the Moles Withdrawn

Now we can find the moles of oxygen that were withdrawn:

$$\text{Moles withdrawn} = n_i - n_{\text{final}} = 18.20 - 14.544 = 3.656 \, \text{moles}$$

Step 4: Calculate the Mass Withdrawn

To find the mass of the oxygen withdrawn, we use the molar mass of oxygen ($O_2$), which is $32 \, \text{g/mol}$ or $0.032 \, \text{kg/mol}$:

$$\text{Mass withdrawn} = \text{Moles withdrawn} \times \text{Molar mass} = 3.656 \, \text{moles} \times 0.032 \, \text{kg/mol}$$

Calculating that:

$$\text{Mass withdrawn} = 3.656 \times 0.032 = 0.116 \, \text{kg}$$

Thus, the mass of the oxygen withdrawn from the cylinder is nearly equal to 0.116 kg.

Conclusion

The correct answer is Option C: 0.116 kg.

Explanation of Other Options

1. Option A (0.125 kg): This value is higher than the calculated mass and results from incorrect calculations or assumptions.
2. Option B (0.144 kg): This is also higher than the calculated value, likely due to a miscalculation in pressure or volume.
3. Option D (0.156 kg): This is the highest option and is not supported by the calculations performed.

In summary, through the application of the ideal gas law and conversion