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physics ncert solutions class 11th Physics Class 11th Kinetic Theory

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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J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

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Payal Gupta | Contributor-Level 10

5 months ago

13.4 Volume of oxygen, $V_{1}$ = 30 litres = 30 $* 10^{- 3} m^{3}$

Gauge pressure, $P_{1}$ = 15 atm = 15 $* 1.013 * 10^{5}$ Pa

Temperature, $T_{1}$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_{1}$

From the gas equation, we get $P_{1} V_{1} = n_{1} R T_{1}$

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ = $\frac{15 * 1.013 * 10^{5} * 30 * 10^{- 3}}{8.314 * 300}$ = 18.276

But $n_{1}$ = $\frac{m_{1}}{M}$ , where $m_{1}$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{1}$ = $n_{1}$ M = 18.276 $* 32 = 584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_{2}$ = 30 litres = 30

...more

13.4 Volume of oxygen, $V_{1}$ = 30 litres = 30 $* 10^{- 3} m^{3}$

Gauge pressure, $P_{1}$ = 15 atm = 15 $* 1.013 * 10^{5}$ Pa

Temperature, $T_{1}$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_{1}$

From the gas equation, we get $P_{1} V_{1} = n_{1} R T_{1}$

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ = $\frac{15 * 1.013 * 10^{5} * 30 * 10^{- 3}}{8.314 * 300}$ = 18.276

But $n_{1}$ = $\frac{m_{1}}{M}$ , where $m_{1}$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{1}$ = $n_{1}$ M = 18.276 $* 32 = 584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_{2}$ = 30 litres = 30 $* 10^{- 3} m^{3}$

Gauge pressure, $P_{2}$ = 11 atm = 11 $* 1.013 * 10^{5}$ Pa

Temperature, $T_{2}$ = 17 $?$ = 290 K

Let the final number of moles of oxygen left in the cylinder be $n_{2}$

From the gas equation, we get $P_{2} V_{2} = n_{2} R T_{2}$

$n_{2} = \frac{P_{2} V_{2}}{R 2}$ = $\frac{11 * 1.013 * 10^{5} * 30 * 10^{- 3}}{8.314 * 290}$ = 13.864

But $n_{2}$ = $\frac{m_{2}}{M}$ , where $m_{2}$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{2}$ = $n_{2}$ M = 13.864 $* 32 = 443.65$ g

So the mass of oxygen taken out = $m_{1} - m_{2}$ = 141.19 gm = 0.141 kg

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