States of Matter

To determine the pressure ratio of the two gases, we can use the ideal gas law given by

$PV = nRT$

where:

• $ P $ = Pressure
• $ V $ = Volume
• $ n $ = Number of moles
• $ R $ = Universal gas constant
• $ T $ = Temperature

Since the volumes and temperatures of both gases are the same, we can simplify the relationship to

$P \propto n$

Therefore, the pressure ratio will be equal to the ratio of the number of moles of the two gases. First, we need to find the number of moles of each gas.

The number of moles is calculated as:

$n = \frac{m}{M}$

where $ m $ is the mass and $ M $ is the molar mass.

For methane, $ \text{CH}_4 $:

$M_{\mathrm{CH}_4} = 12 + 4 \times 1 = 16 \mathrm{\ g/mol}$

$n_{\mathrm{CH}_4} = \frac{10 \mathrm{\ g}}{16 \mathrm{\ g/mol}} = 0.625 \mathrm{\ mol}$

For oxygen, $ \mathrm{O}_2 $:

$M_{\mathrm{O}_2} = 2 \times 16 = 32 \mathrm{\ g/mol}$

$n_{\mathrm{O}_2} = \frac{10 \mathrm{\ g}}{32 \mathrm{\ g/mol}} = 0.3125 \mathrm{\ mol}$

Now, we calculate the pressure ratio:

$\frac{P_{\mathrm{CH}_4}}{P_{\mathrm{O}_2}} = \frac{n_{\mathrm{CH}_4}}{n_{\mathrm{O}_2}} = \frac{0.625}{0.3125} = 2$

Therefore, the pressure ratio of $ \mathrm{CH}_4 $ to $ \mathrm{O}_2 $ is 2:1.

The correct option is Option D.

The correct order of pressure is $p_1>p_3>p_2$.

Option C In case of liquids, decreases and in case of gases, increases.

Explanation :

Viscosity, in general, refers to a fluid's resistance to flow. It can be thought of as the internal friction between molecules in a fluid (liquids or gases).

For liquids, as temperature increases, the viscosity usually decreases. This is because at higher temperatures, the energy in the liquid is greater, allowing particles to move more easily and reducing the internal friction (or viscosity).

In contrast, for gases, as temperature increases, viscosity tends to increase. This is due to the fact that, in gases, increased temperatures lead to higher molecular speeds and more frequent collisions, effectively increasing the internal friction or viscosity.

From the graph it is clear that, the value of 'Z' decreases with increase of pressure. We can explain as follows on the basis of van der Waals' equation.

At high pressure, when 'p' is large, V will be small and one cannot ignore 'b' in comparison to V. However, the team a/V² may be considered negligible in comparison to 'p' in van der Waals' equation.

$\left(p+\frac{a}{V^2}\right)$ (V $-$ b) = nRT

p(V $-$ b) = nRT

$\Rightarrow$ pV $-$ pb = nRT

or $\frac{pV}{nRT}=1+\frac{pb}{nRT}$

or Z = 1 + $\frac{pb}{nRT}$

Thus, Z is greater than 1. As pressure is increased (at constant T), the factor $\frac{pb}{nRT}$ increases. This explains why after minima in the curves, Z increase continuously with pressure. Hence, the only incorrect statement is (b).

Suppose the equal mass of methane and oxygen, w = 1 g

Mole fraction of oxygen = $ = {{w/32} \over {{w \over {32}} + {w \over {16}}}} = {{1/32} \over {{3 \over {32}}}} = {1 \over 3}$

Let, the total pressure = p

Pressure exerted by oxygen (partial pressure)

$\therefore$ The fraction of total pressure exerted by oxygen is ${1 \over 3}p$.

Statement I and III are correct and statement II incorrect.

(I) ${{{T_c}} \over {{p_c}}} = {{8a} \over {27\,Rb}}.\,{{27{b^2}} \over a} = {{8b} \over R}$ $\therefore$ ${{{T_c}} \over {{p_c}}} \propto b$

(II) ${T_c} = {{8a} \over {27\,Rb}}$ and ${T_b} = {a \over {Rb}}$

Thus, T_b > T_c

(III) When, T_c is attained.

$\therefore$ $\left( {{{\partial p} \over {\partial V}}} \right)$ = 0