In given figure shows plot of \( \dfrac{PV}{T}\) versus P for \( 1.00×10^{-3}\) kg of oxygen gas at two different temperatures.

**(a)** What does the dotted plot signify?

**(b)** Which is true: T1 > T2 or T1 < T2?

**(c)** What is the value of \( \dfrac{PV}{T}\) where the curves meet on the y-axis?

**(d)** If we obtained similar plots for 1.00×10-3 kg of hydrogen, would we get the same value of \( \dfrac{PV}{T}\) at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of \( \dfrac{PV}{T}\)(for low pressure high temperature region of the plot)? (\( Molecular;\ mass\; of\; H_2 = 2.02 u, of \)

\( O_2 = 32.0 u, R = 8.31 J mo1^{–1} K{–1})\)

Asked by Pragya Singh | 1 year ago | 148

**(a)** dotted plot is parallel to X-axis, signifying that nR [\( \dfrac{PV}{T}\) = nR] is independent of P. Thus it is representing ideal gas behaviour

**(b)** the graph at temperature T_{1} is closer to ideal behaviour (because closer to dotted line) hence, T_{1} > T_{2} (higher the temperature, ideal behaviour is the higher)

**(c)** use PV = nRT

\( \dfrac{PV}{T}\) = nR

Mass of the gas = 1 x 10^{-3} kg = 1 g

Molecular mass of O_{2} = 32g/mol

Hence,

Number of mole = given weight/molecular weight

= \( \dfrac{1}{32}\)

So, nR = \( \dfrac{1}{32}\) x 8.314 = 0.26 J/K

Hence,

Value of \( \dfrac{PV}{T}\)= 0.26 J/K

**(d)** 1 g of H_{2} doesn’t represent the same number of mole

Eg. molecular mass of H_{2} = 2 g/mol

Hence, number of moles of H_{2} require is \( \dfrac{1}{32}\) (as per the question)

Therefore,

Mass of H_{2} required = no. of mole of H_{2} x molecular mass of H_{2}

= \( \dfrac{1}{32}\) x 2

= \( \dfrac{1}{16}\) g

= 0.0625 g

= 6.3 x 10^{-5} kg

Hence, 6.3 x 10^{-5} kg of H_{2} would yield the same value

Given 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).