Physics Q14 – NEET-UG 2025

Home Physics ThermodynamicsQuestion #517

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to : [Given,

R = \frac{100}{12} {J mol}^{- 1} K^{- 1}

, and molecular mass of

O_{2} = 32

, 1 atm pressure =

1.01 \times 10^{5} N/m

]

Clue 1 of 3: Hint

🥷

Ninja Strategy•Proportionality Estimation

Quickly estimate the initial pressure (~15 atm). Since moles are proportional to pressure, the fractional drop in moles is $(15 - 12) / 15 = 1 / 5$ . The mass withdrawn is about 1/5 of the initial mass ( $18.2 \times 32 g$ ), which is approximately 0.116 kg.

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● Video Walkthrough

Step 1: Determine the Final State of the Gas in SI Units✦ Active

First, convert all given quantities to standard SI units. The temperature is $T = 27^{\circ} C = 27 + 273 = 300 K$ . The volume is $V = 30 litre = 30 \times 10^{- 3} m^{3}$ . The final pressure is given as gauge pressure, so we must find the absolute pressure. Assuming atmospheric pressure is 1 atm, the final absolute pressure is $P_{f i n a l} = P_{g a u g e} + P_{a t m} = 11 atm + 1 atm = 12 atm$ . Convert this to Pascals (N/m²) using the given conversion: $P_{f i n a l} = 12 \times (1.01 \times 10^{5} {N/m}^{2}) = 12.12 \times 10^{5} Pa$ .

💡 Teacher's Secret Hint

Note: The unit for pressure in the problem is given as N/m, which is likely a typo for N/m² (Pascals). The calculation proceeds assuming the standard unit for pressure.

Step 2: Calculate the Final Number of Moles○ Expand

Use the Ideal Gas Law, $P V = n R T$ , to find the number of moles of oxygen remaining in the cylinder, $n_{2}$ . Rearranging the formula gives $n_{2} = \frac{P_{f i n a l} V}{R T}$ .

n_{2} = \frac{(12.12 \times 10^{5} {N/m}^{2}) \times (30 \times 10^{- 3} m^{3})}{(\frac{100}{12} {J mol}^{- 1} K^{- 1}) \times (300 K)}

Solving this gives: $n_{2} = \frac{36360}{2500} = 14.544 mol$ .

Step 3: Calculate the Mass of Withdrawn Oxygen○ Expand

The number of moles of oxygen withdrawn is the difference between the initial and final moles: $Δ n = n_{i n i t i a l} - n_{f i n a l} = 18.20 - 14.544 = 3.656 mol$ . The mass withdrawn is $Δ m = Δ n \times M$ , where $M$ is the molecular mass of $O_{2}$ ( $32 g/mol$ or $0.032 kg/mol$ ).

Δ m = 3.656 mol \times 0.032 kg/mol = 0.116992 kg

This value is approximately $0.117$ kg. Comparing with the options, the closest value is $0.116$ kg.

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Physics Question 14 – NEET-UG 2025

Clue 1 of 3: Hint

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