Question

1/ An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to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.
2/ A balloon whose volume is 750 m3 is to be filled with hydrogen at atmospheric pressure
(1.01105 Pa).
(a) If the hydrogen is stored in cylinders with volumes of 1.90 m3 at a gauge pressure of 1.20  106 Pa, how many cylinders are required? Assume that the temperature of the hydrogen remains constant.
(b) What is the total weight (in addition to the weight of the gas) that can be supported by the
balloon if the gas in the balloon and the surrounding air are both at 15.0°C? The density of air at
15.0oC and atmospheric pressure is 1.23 kg/m3.
(c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 g/mol) instead of hydrogen at 15.0°C? What are your observations?

Answer 1

Answer on Question #56494-Physics-Other

1/ An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of $27{}^{\circ}\mathrm{C}$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to $17{}^{\circ}\mathrm{C}$. Estimate the mass of oxygen taken out of the cylinder.

Solution

$$(R = 8.31 \, \mathrm{Jmol}^{-1} \, K^{-1}, \text{ molecular mass of } O_2 = 32 \, \mu).$$

Volume of oxygen, $V_1 = 30 \, \mathrm{litres} = 30 \cdot 10^{-3} \, \mathrm{m}^3$.

Gauge pressure, $P_1 = 15 \, \mathrm{atm} = 15 \cdot 1.013 \cdot 10^5 \, \mathrm{Pa}$.

Temperature, $T_1 = 27{}^{\circ}C = 300 \, K$.

Universal gas constant, $R = 8.31 \, \mathrm{Jmol}^{-1} \, K^{-1}$.

Let the initial number of moles of oxygen gas in the cylinder be $n_1$.

The gas equation is given as:

$$P_1 V_1 = n_1 R T_1 \rightarrow n_1 = \frac{P_1 V_1}{R T_1}$$ $$n_1 = \frac{(15.195 \cdot 10^5 \cdot 30 \cdot 10^{-3})}{(8.314 \cdot 300)} = 18.276.$$

But $n_1 = \frac{m_1}{M}$.

Where, $m_1$ is Initial mass of oxygen, $M = 32 \, \mathrm{g}$ is Molecular mass of oxygen. So,

$$m_1 = n_1 M = 18.276 \cdot 32 = 584.84 \, \mathrm{g}$$

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, $V_2 = 30 \, \mathrm{litres} = 30 \cdot 10^{-3} \, \mathrm{m}^3$.

Gauge pressure, $P_2 = 11 \, \mathrm{atm} = 11 \cdot 1.013 \cdot 10^5 \, \mathrm{Pa}$.

Temperature, $T_2 = 17{}^{\circ}C = 290 \, K$.

Let $n_2$ be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

$$P_2 V_2 = n_2 R T_2 \rightarrow n_2 = \frac{P_2 V_2}{R T_2}$$ $$n_2 = \frac{(11.143 \cdot 10^5 \cdot 30 \cdot 10^{-3})}{(8.314 \cdot 290)} = 13.86.$$

But $n_2 = \frac{m_2}{M}$.

Where, $m_2$ is the mass of oxygen remaining in the cylinder.

$$m _ {2} = n _ {2} M = 1 3. 8 6 \cdot 3 2 = 4 5 3. 1 g.$$

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

$$\begin{array}{l} \text{Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder} = m _ {1} - m _ {2} \\ = 5 8 4. 8 4 \mathrm {g} - 4 5 3. 1 \mathrm {g} = 1 3 1. 7 4 \mathrm {g} = 0. 1 3 1 \mathrm {k g}. \\ \end{array}$$

Therefore, $0.131\mathrm{kg}$ of oxygen is taken out of the cylinder.

2/ A balloon whose volume is $750\mathrm{m}^3$ is to be filled with hydrogen at atmospheric pressure (1.01@105 Pa).

(a) If the hydrogen is stored in cylinders with volumes of $1.90\mathrm{m}^3$ at a gauge pressure of 1.20 @ 106 Pa, how many cylinders are required? Assume that the temperature of the hydrogen remains constant.

(b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at $15.0{}^{\circ}\mathrm{C}$? The density of air at 15.0oC and atmospheric pressure is $1.23\mathrm{kg / m}^3$.

(c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 g/mol) instead of hydrogen at $15.0{}^{\circ}\mathrm{C}$? What are your observations?

Solution

(a) The absolute pressure of the gas in a cylinder is:

$$P = 1. 2 0 \cdot 1 0 ^ {6} + 1. 0 1 \cdot 1 0 ^ {5} = 1. 3 0 \cdot 1 0 ^ {6} P a.$$

At atmospheric pressure, the volume of hydrogen will increase by a factor of

$$\frac {1. 3 0 \cdot 1 0 ^ {6}}{1. 0 1 \cdot 1 0 ^ {5}}$$

so the number of cylinders is:

$$N = \frac {7 5 0 m ^ {3}}{1. 9 0 m ^ {3} \left(\frac {1 . 3 0 \cdot 1 0 ^ {6}}{1 . 0 1 \cdot 1 0 ^ {5}}\right)} = 3 1.$$

(b) The difference between the weight of the air displaced and the weight of hydrogen is:

$$\begin{array}{l} W _ {H _ {2}} = \left(\rho_ {\text {air}} - \rho_ {H _ {2}}\right) V g = \left(\rho_ {\text {air}} - \frac {p M _ {H _ {2}}}{R T}\right) V g \\ = \left(1. 2 3 \frac {k g}{m ^ {3}} - \frac {1. 0 1 \cdot 1 0 ^ {5} P a \cdot 2 . 2 \cdot 1 0 ^ {- 3} \frac {k g}{m o l}}{8 . 3 1 4 \frac {J}{m o l K} \cdot 2 8 8 . 1 5 K}\right) 9. 8 0 \frac {m}{s ^ {2}} \cdot 7 5 0 m ^ {3} = 8. 4 2 \cdot 1 0 ^ {3} N \\ = 8. 4 2 k N. \\ \end{array}$$

(c) The difference between the weight of the air displaced and the weight of helium is:

$$W_{He_2} = (\rho_{air} - \rho_{He_2}) V g = \left(\rho_{air} - \frac{p M_{He_2}}{R T}\right) V g$$ $$= \left(1.23 \frac{k g}{m^3} - \frac{1.01 \cdot 10^5 P a \cdot 4.00 \cdot 10^{-3} \frac{k g}{m o l}}{8.314 \frac{J}{m o l K} \cdot 288.15 K}\right) 9.80 \frac{m}{s^2} \cdot 750 m^3 = 7.80 \cdot 10^3 N$$ $$= 7.80 \, kN.$$

It is less than for hydrogen.

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