# An Oxygen Cylinder of Volume 30 Litres Has an Initial Gauge Pressure of 15 Atm and a Temperature of 27 °C. Estimate the Mass of Oxygen Taken Out of the Cylinder - Physics

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (= 8.31 J mol–1 K–1, molecular mass of O2 = 32 u)

#### Solution 1

Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3

Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa

Temperature, T1 = 27°C = 300 K

Universal gas constant, R = 8.314 J mole–1 K–1

Let the initial number of moles of oxygen gas in the cylinder be n1.

The gas equation is given as:

P_1V_1 = eta_1RT_1

:.n_1 = P_1V_1/RT_1

= (15.195 xx10^5 xx 30 xx 10^(-3))/((8.314) xx 300) = 18.276

Where,

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

m1 = n1= 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, V2 = 30 litres = 30 × 10–3 m3

Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa

Temperature, T2 = 17°C = 290 K

Let n2 be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

P2V2 = n2RT2

:. n_2= (P_2V_2)/(RT_2)

= (11.143 xx 10^5 xx 30 xx 10^(-3))/8.314 xx 290 = 13.86

But n_2 = m_2/M

Where,

m2 is the mass of oxygen remaining in the cylinder

m2 = n2= 13.86 × 32 = 453.1 g

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

m1 – m2

= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

#### Solution 2

Initial Volume, V_1 = 30 litre = 30 xx 10^3 cm^3

= 30 xx 10^3 xx 10^(-6) m^3 = 30 xx 10^(-3) m^3

Initla Pressure, P_1 = 15 atm

= 15 xx1.013 xx 10^5 N m^(-2)

Inital temperature, T_1 = (27 + 273) K = 300 K

Initial number of moles

mu_1 = (P_1V_1)/RT_1 = (15xx1.013 xx10^5 xx 30xx10^(-3))/(8.31 xx 300) = 18.3

Final Pressure, P_2 = 11 atm

= 11 xx 1.013 xx 10^5 N m^(-2)

Final Volume, V_2 = 30 litre = 30 xx 10^(-3) cm^3

Final temperature, T_2 = 17 + 273 = 290 K

Final number of moless, mu_2 = (P_2V_2)/(RT_2) = (11xx 1.013xx 10^5xx 30xx10^(-3))/(8.31 xx 290) = 13.9

Number of moles takesout of cyclinder = 18.3 - 13.9 = 4.4

Mass of gas taken out of cylinder = 4.4 xx 32 g = 140.8 g = 0.141 kg

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 13 Kinetic Theory
Q 4 | Page 334