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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (\( R = 8.31 J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 \( O_2 = 32 u\)).

Asked by Pragya Singh | 1 year ago | 122

Volume of gas, V_{1} = 30 litres

= 30 x 10^{-3 }m^{3}

Gauge pressure, P_{1} = 15 atm

= 15 x 1.013 x 10^{5} P a

Temperature, T_{1} = 27°C

= 300 K

Universal gas constant, R = 8.314 J mol^{-1} K^{-1}

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

The gas equation is given as follows:

P_{1}V_{1} = n_{1}RT_{1}

Hence,

n_{1} = \( \dfrac{P_1V_1}{RT_1}\)

= \( \dfrac{(15.195 \times 10^5 \times 30 \times 10^{-3)}}{(8.314 \times 300)}\)

= 18.276

But n_{1} = \( \dfrac{m_1}{M}\)

Where,

m_{1} = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Thus,

m_{1} = N_{1}M = 18.276 x 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.

Volume, V_{2} = 30 litres = 30 x 10^{-3} m^{3}

Gauge pressure, P_{2} = 11 atm

= 11 x 1.013 x 10^{5 }P a

Temperature, T_{2} = 17°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}RT_{2}

Hence,

n_{2} = \( \dfrac{P_2V_2}{RT_2}\)

= \( \dfrac{(11.143 \times 10^5 \times 30 \times 10^{-30)}}{ (8.314 \times 290)}\)

= 13.86

But

n_{2} = \( \dfrac{m_2}{M}\)

Where,

m_{2 }is the mass of oxygen remaining in the cylinder

Therefore,

m_{2} = n_{2} x M = 13.86 x 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

= m_{1} – m_{2}

= 584.84 g – 453.1 g

We get,

= 131.74 g

= 0.131 kg

Hence, 0.131 kg of oxygen is taken out of the cylinder

Answered by Pragya Singh | 1 year agoGiven below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance |
Atomic Mass (u) |
Kg m^{-3}) |

Carbon (diamond) | 12.01 | 2.22 |

Gold | 197.00 | 19.32 |

Nitrogen (liquid) | 14.01 | 1.00 |

Lithium | 6.94 | 0.53 |

Fluorine (liquid) | 19.00 | 1.14 |

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n_{2}= n_{1} exp [ -mg \(
\dfrac{(h_2–h_1)}{k_BT}\)T] where n_{2}, n_{1 }refer to number density at heights h_{2 }and h_{1 }respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:n_{2} = n_{1} exp [ -mg N_{A} (ρ – ρ′ ) \(
\dfrac{(h_2–h_1)}{ρ RT}\)] where ρ is the density of the suspended particle, and ρ′ ,that of surrounding medium. [N_{A} is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

From a certain apparatus, the diffusion rate of hydrogen has an average value of \( 28.7 cm^3 s^{-1}\). The diffusion of another gas under the same conditions is measured to have an average rate of \( 7.2 cm^3 s^{-1}\). Identify the gas. [Hint: Use Graham’s law of diffusion: \( \dfrac{R_1}{R_2}\) = \( (\dfrac{M_2}{M_1})^ \dfrac{1}{2}\), where \( R_1 , R_2\) are diffusion rates of gases 1 and 2, and \( M_1\) and \( M_2\) their respective molecular masses. The law is a simple consequence of kinetic theory.]

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 170°C . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of \( N_2\) = 28.0 u).