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Q12.4: NCERT Physics - Part 2 KINETIC THEORY 12.4 Solution: An oxygen cylinder of volume 30 litre has an initial gauge…

An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 �C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 �C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol__ K__, molecular mass of O_ = 32 u).

Step-by-Step Verified Solution

**Answer:** Mass of oxygen taken out = 0.14 kg **Step-by-Step Explanation:** Volume of cylinder V=30V = 30 L =30×10−3= 30 \times 10^{-3} m_. **Initial state:** Gauge pressure P1=15P_1 = 15 atm, so absolute pressure P1,abs=15+1=16P_{1,abs} = 15 + 1 = 16 atm =16×1.013×105= 16 \times 1.013 \times 10^5 Pa. T1=27∘C=300T_1 = 27^\circ\text{C} = 300 K. Number of moles initially:
n1=P1,absVRT1=16×1.013×105×30×10−38.31×300=19.5 moln_1 = \frac{P_{1,abs}V}{RT_1} = \frac{16 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 300} = 19.5 \text{ mol}
**Final state:** Gauge pressure P2=11P_2 = 11 atm, so absolute P2,abs=12P_{2,abs} = 12 atm =12×1.013×105= 12 \times 1.013 \times 10^5 Pa. T2=17∘C=290T_2 = 17^\circ\text{C} = 290 K. Number of moles finally:
n2=P2,absVRT2=12×1.013×105×30×10−38.31×290=15.1 moln_2 = \frac{P_{2,abs}V}{RT_2} = \frac{12 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 290} = 15.1 \text{ mol}
Moles withdrawn: Δn=n1−n2=19.5−15.1=4.4\Delta n = n_1 - n_2 = 19.5 - 15.1 = 4.4 mol. Mass withdrawn:
Δm=Δn×M=4.4×32=140.8 g=0.141 kg\Delta m = \Delta n \times M = 4.4 \times 32 = 140.8 \text{ g} = 0.141 \text{ kg}

All Questions in this Chapter

Q12.1 Q12.2 Q12.3 Q12.4 Q12.5 Q12.6 Q12.7 Q12.8 Q12.9 Q12.10

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