NCERT Solutions For Class 11 Chemistry chapter 5- States Of Matter(Outdated)

Question 1:

What minimum pressure is required to compress $500\text{ dm}^3$ of air at $1\text{ bar}$ to $200\text{ dm}^3$ at $30^\circ\text{C}$?

Solution:

Given,
Initial pressure, $P_1 = 1\text{ bar}$
Initial volume, $V_1 = 500\text{ dm}^3$
Final volume, $V_2 = 200\text{ dm}^3$
Since the temperature remains constant, the final pressure ($P_2$) can be calculated using Boyle’s law.
According to Boyle’s law,
$$P_1 V_1 = P_2 V_2$$
$$P_2 = \frac{P_1 V_1}{V_2} = \frac{1\text{ bar} \times 500\text{ dm}^3}{200\text{ dm}^3} = 2.5\text{ bar}$$
Minimum required pressure $= 2.5\text{ bar}$

Question 2:

A vessel of $120\text{ mL}$ capacity contains a certain amount of gas at $35^\circ\text{C}$ and $1.2\text{ bar}$ pressure. The gas is transferred to another vessel of volume $180\text{ mL}$ at $35^\circ\text{C}$. What would be its pressure?

Solution:

Initial pressure, $P_1 = 1.2\text{ bar}$
Initial volume, $V_1 = 120\text{ mL}$
Final volume, $V_2 = 180\text{ mL}$
As the temperature remains same, final pressure ($P_2$) can be calculated with the help of Boyle’s law.
According to the Boyle’s law,
$$P_1 V_1 = P_2 V_2$$
$$P_2 = \frac{1.2\text{ bar} \times 120\text{ mL}}{180\text{ mL}} = 0.80\text{ bar}$$
Final pressure $= 0.80\text{ bar}$

Question 3:

Using the equation of state $PV = nRT$, show that at a given temperature the density of a gas is proportional to its pressure.

Solution:

The equation of state is given by,
$$pV = nRT \quad \dots\text{(i)}$$
Where,
$p \rightarrow \text{Pressure of gas}$
$V \rightarrow \text{Volume of gas}$
$n \rightarrow \text{Number of moles of gas}$
$R \rightarrow \text{Gas constant}$
$T \rightarrow \text{Temperature of gas}$

From equation (i) we have,
$$p = \frac{n}{V} RT \quad \dots\text{(ii)}$$
Replacing $n$ with $\frac{m}{M}$, we have:
$$p = \frac{m}{V} \frac{RT}{M}$$
Where,
$m \rightarrow \text{Mass of gas}$
$M \rightarrow \text{Molar mass of gas}$
But, $\frac{m}{V} = d$ ($d = \text{density of gas}$)
Thus, from equation (ii), we have:
$$p = d \frac{RT}{M} \implies d = \left(\frac{M}{RT}\right)p$$
Molar mass ($M$) of a gas is always constant and therefore, at constant temperature $\frac{M}{RT} = \text{constant}$.
Hence, at a given temperature, the density ($d$) of gas is proportional to its pressure ($p$).

Question 4:

At $0^\circ\text{C}$, the density of a certain oxide of a gas at $2\text{ bar}$ is same as that of dinitrogen at $5\text{ bar}$. What is the molecular mass of the oxide?

Solution:

Density ($d$) of the substance at temperature ($T$) can be given by the expression,
$$d = \frac{pM}{RT}$$
Now, density of oxide ($d_1$) is given by,
$$d_1 = \frac{p_1 M_1}{RT}$$
Where, $M_1$ and $p_1$ are the mass and pressure of the oxide respectively.
Density of dinitrogen gas ($d_2$) is given by,
$$d_2 = \frac{p_2 M_2}{RT}$$
Where, $M_2$ and $p_2$ are the mass and pressure of dinitrogen respectively.
According to the given question, $d_1 = d_2$:
$$\frac{p_1 M_1}{RT} = \frac{p_2 M_2}{RT} \implies p_1 M_1 = p_2 M_2$$
Given, $p_1 = 2\text{ bar}$, $p_2 = 5\text{ bar}$, Molecular mass of nitrogen, $M_2 = 28\text{ g/mol}$:
$$2 \times M_1 = 5 \times 28 \implies M_1 = \frac{140}{2} = 70\text{ g/mol}$$
Hence, the molecular mass of the oxide is $70\text{ g/mol}$.

Question 5:

The pressure of $1\text{ g}$ of an ideal gas A at $27^\circ\text{C}$ is $2\text{ bar}$. When $2\text{ g}$ of another ideal gas B is introduced into the same flask at the same temperature, the total pressure becomes $3\text{ bar}$. Find the relationship between their molar masses.

Solution:

Suppose molecular masses of A and B are $M_A$ and $M_B$ respectively. Then their number of moles will be $n_A = \frac{1}{M_A}$ and $n_B = \frac{2}{M_B}$.
Using the ideal gas equation:
$$p_A V = n_A R T \implies 2 V = \left(\frac{1}{M_A}\right) R T$$
Total pressure after introducing B:
$$p_{\text{total}} V = (n_A + n_B) R T \implies 3 V = \left(\frac{1}{M_A} + \frac{2}{M_B}\right) R T$$
Dividing the equations:
$$\frac{2}{3} = \frac{1/M_A}{1/M_A + 2/M_B} \implies \frac{2}{M_A} + \frac{4}{M_B} = \frac{3}{M_A} \implies \frac{1}{M_A} = \frac{4}{M_B} \implies M_B = 4 M_A$$

Question 6:

The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20^\circ\text{C}$ and one bar will be released when $0.15\text{ g}$ of aluminum reacts?

Solution:

The reaction of aluminium with caustic soda can be represented as:
$$2\text{Al} + 2\text{NaOH} + 2\text{H}_2\text{O} \rightarrow 2\text{NaAlO}_2 + 3\text{H}_2$$
At STP ($273.15\text{ K}$ and $1\text{ atm}$), $54\text{ g}$ ($2 \times 27\text{ g}$) of Al gives $3 \times 22400\text{ mL}$ of $\text{H}_2$.
$0.15\text{ g}$ Al gives:
$$\frac{3 \times 22400 \times 0.15}{54} = 186.67\text{ mL of }\text{H}_2\text{ at STP}$$
At STP: $P_1 = 1\text{ atm}$, $V_1 = 186.67\text{ mL}$, $T_1 = 273.15\text{ K}$
Let the volume of dihydrogen be at $P_2 = 0.987\text{ atm}$ (since $1\text{ bar} = 0.987\text{ atm}$) and $T_2 = 20^\circ\text{C} = (273.15 + 20)\text{ K} = 293.15\text{ K}$.
Using the combined gas law:
$$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{1 \times 186.67 \times 293.15}{273.15 \times 0.987} \approx 203\text{ mL}$$
Therefore, $203\text{ mL}$ of dihydrogen will be released.

Question 7:

What will be the pressure exerted by a mixture of $3.2\text{ g}$ of methane and $4.4\text{ g}$ of carbon dioxide contained in a $9\text{ dm}^3$ flask at $27^\circ\text{C}$?

Solution:

It is known that, $PV = nRT$
For methane ($\text{CH}_4$, molar mass $= 16\text{ g/mol}$):
$$n_{\text{CH}_4} = \frac{3.2\text{ g}}{16\text{ g/mol}} = 0.2\text{ mol}$$
For carbon dioxide ($\text{CO}_2$, molar mass $= 44\text{ g/mol}$):
$$n_{\text{CO}_2} = \frac{4.4\text{ g}}{44\text{ g/mol}} = 0.1\text{ mol}$$
Total moles $n = 0.2 + 0.1 = 0.3\text{ mol}$
Volume $V = 9\text{ dm}^3 = 9 \times 10^{-3}\text{ m}^3$, Temperature $T = 27^\circ\text{C} = 300\text{ K}$, $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$
Total pressure exerted by the mixture can be obtained as:
$$P = \frac{nRT}{V} = \frac{0.3 \times 8.314 \times 300}{9 \times 10^{-3}} = 8.314 \times 10^4\text{ Pa}$$
Hence, the total pressure exerted by the mixture is $8.314 \times 10^4\text{ Pa}$.

Question 8:

What will be the pressure of the gaseous mixture when $0.5\text{ L}$ of $\text{H}_2$ at $0.8\text{ bar}$ and $2.0\text{ L}$ of dioxygen at $0.7\text{ bar}$ are introduced in a $1\text{ L}$ vessel at $27^\circ\text{C}$?

Solution:

Calculation of partial pressure of $\text{H}_2$ in $1\text{ L}$ vessel ($P_{\text{H}_2}$):
$P_1 = 0.8\text{ bar}, V_1 = 0.5\text{ L}, V_2 = 1.0\text{ L}$
As temperature remains constant, $P_1 V_1 = P_2 V_2$:
$$(0.8\text{ bar})(0.5\text{ L}) = P_{\text{H}_2} (1.0\text{ L}) \implies P_{\text{H}_2} = 0.40\text{ bar}$$
Calculation of partial pressure of $\text{O}_2$ in $1\text{ L}$ vessel ($P_{\text{O}_2}$):
$P_1′ = 0.7\text{ bar}, V_1′ = 2.0\text{ L}, V_2′ = 1.0\text{ L}$:
$$(0.7\text{ bar})(2.0\text{ L}) = P_{\text{O}_2} (1.0\text{ L}) \implies P_{\text{O}_2} = 1.4\text{ bar}$$
Total pressure $= P_{\text{H}_2} + P_{\text{O}_2} = 0.4\text{ bar} + 1.4\text{ bar} = 1.8\text{ bar}$.

Question 9:

Density of a gas is found to be $5.46\text{ g/dm}^3$ at $27^\circ\text{C}$ at $2\text{ bar}$ pressure. What will be its density at STP?

Solution:

Using $\frac{d_1 T_1}{P_1} = \frac{d_2 T_2}{P_2}$:
$$d_1 = 5.46\text{ g/dm}^3, T_1 = 300\text{ K}, P_1 = 2\text{ bar}$$
$$T_2 = 273.15\text{ K}, P_2 = 1.013\text{ bar}$$
$$d_2 = \frac{d_1 P_2 T_1}{P_1 T_2} = \frac{5.46 \times 1.013 \times 300}{2 \times 273.15} \approx 3.00\text{ g dm}^{-3}$$
Density at STP $\approx 3.00\text{ g dm}^{-3}$.

Question 10:

$34.05\text{ mL}$ of phosphorus vapour weighs $0.0625\text{ g}$ at $546^\circ\text{C}$ and $0.1\text{ bar}$ pressure. What is the molar mass of phosphorus?

Solution:

Given,
$p = 0.1\text{ bar}$
$V = 34.05\text{ mL} = 34.05 \times 10^{-3}\text{ L} = 34.05 \times 10^{-3}\text{ dm}^3$
$R = 0.083\text{ bar dm}^3\text{ K}^{-1}\text{ mol}^{-1}$
$T = 546^\circ\text{C} = (546 + 273)\text{ K} = 819\text{ K}$
$w = 0.0625\text{ g}$
The number of moles ($n$) can be calculated using the ideal gas equation as:
$$pV = nRT = \frac{w}{M} RT \implies M = \frac{wRT}{pV} = \frac{0.0625 \times 0.083 \times 819}{0.1 \times 34.05 \times 10^{-3}} = 1247.5\text{ g mol}^{-1}$$
Hence, the molar mass of phosphorus is $1247.5\text{ g mol}^{-1}$.

Question 11:

A student forgot to add the reaction mixture to the round bottomed flask at $27^\circ\text{C}$ but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was $477^\circ\text{C}$. What fraction of air would have been expelled out?

Solution:

Initial temperature $T_1 = 27^\circ\text{C} = 300\text{ K}$
Final temperature $T_2 = 477^\circ\text{C} = 750\text{ K}$
Since pressure and volume remain constant, $n_1 T_1 = n_2 T_2$:
$$\frac{n_2}{n_1} = \frac{T_1}{T_2} = \frac{300}{750} = \frac{2}{5} = 0.4$$
Fraction of air remaining $= 0.4$.
Fraction of air expelled out $= 1 – 0.4 = 0.6$ (or $60\%$).

Question 12:

Calculate the temperature of $4.0\text{ mol}$ of a gas occupying $5\text{ dm}^3$ at $3.32\text{ bar}$ ($R = 0.083\text{ bar dm}^3\text{ K}^{-1}\text{ mol}^{-1}$).

Solution:

Given,
$n = 4.0\text{ mol}$
$V = 5\text{ dm}^3$
$p = 3.32\text{ bar}$
$R = 0.083\text{ bar dm}^3\text{ K}^{-1}\text{ mol}^{-1}$
The temperature ($T$) can be calculated using the ideal gas equation as:
$$T = \frac{pV}{nR} = \frac{3.32 \times 5}{4.0 \times 0.083} = \frac{16.6}{0.332} = 50\text{ K}$$
Hence, the required temperature is $50\text{ K}$.

Question 13:

Calculate the total number of electrons present in $1.4\text{ g}$ of dinitrogen gas.

Solution:

Molar mass of dinitrogen ($\text{N}_2$) $= 28\text{ g mol}^{-1}$
Thus, $1.4\text{ g}$ of $\text{N}_2 = \frac{1.4}{28} = 0.05\text{ mol} = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22}\text{ molecules}$.
Now, $1\text{ molecule}$ of $\text{N}_2$ contains $14\text{ electrons}$.
Therefore, $3.011 \times 10^{22}\text{ molecules}$ of $\text{N}_2$ contains:
$$3.011 \times 10^{22} \times 14 = 4.215 \times 10^{23}\text{ electrons}$$

Question 14:

How much time would it take to distribute one Avogadro number of wheat grains, if $10^{10}\text{ grains}$ are distributed each second?

Solution:

Avogadro number $= 6.022 \times 10^{23}$
Thus, time required:
$$t = \frac{6.022 \times 10^{23}}{10^{10}} = 6.022 \times 10^{13}\text{ seconds} = \frac{6.022 \times 10^{13}}{365 \times 24 \times 3600} \approx 1.9 \times 10^6\text{ years}$$
Hence, the time taken would be $1.9 \times 10^6\text{ years}$.

Question 15:

Calculate the total pressure in a mixture of $8\text{ g}$ of dioxygen and $4\text{ g}$ of dihydrogen confined in a vessel of $1\text{ dm}^3$ at $27^\circ\text{C}$. $R = 0.083\text{ bar dm}^3\text{ K}^{-1}\text{ mol}^{-1}$.

Solution:

Given,
Mass of dioxygen ($\text{O}_2$) $= 8\text{ g} \implies n_{\text{O}_2} = \frac{8}{32} = 0.25\text{ mol}$
Mass of dihydrogen ($\text{H}_2$) $= 4\text{ g} \implies n_{\text{H}_2} = \frac{4}{2} = 2.0\text{ mol}$
Therefore, total number of moles in the mixture $= 0.25 + 2.0 = 2.25\text{ mol}$
Given, $V = 1\text{ dm}^3$, $n = 2.25\text{ mol}$, $R = 0.083\text{ bar dm}^3\text{ K}^{-1}\text{ mol}^{-1}$, $T = 27^\circ\text{C} = 300\text{ K}$
Total pressure ($p$) can be calculated as:
$$pV = nRT \implies p = \frac{nRT}{V} = \frac{2.25 \times 0.083 \times 300}{1} = 56.025\text{ bar}$$
Hence, the total pressure of the mixture is $56.025\text{ bar}$.

Question 16:

Payload is the difference between the mass of displaced air and the total mass of the balloon with its gas. Calculate the payload of a spherical balloon of radius $10\text{ m}$ and mass $100\text{ kg}$, filled with helium at $1.66\text{ bar}$ and $27^\circ\text{C}$. Density of air $= 1.2\text{ kg m}^{-3}$.

Solution:

Given,
Radius of the balloon, $r = 10\text{ m}$
Volume of the balloon $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.1416 \times (10)^3 = 4188.79\text{ m}^3$ (or $4190.5\text{ m}^3$ as standard approx)
Thus, the volume of the displaced air is $4190.5\text{ m}^3$.
Given, Density of air $= 1.2\text{ kg m}^{-3}$
Mass of displaced air $= 4190.5 \times 1.2 = 5028.6\text{ kg}$
Mass of helium ($m$) inside the balloon:
$$m = \frac{PVM}{RT} = \frac{1.66 \times 10^5 \times 4190.5 \times 4}{8.314 \times 300} = 1117.5\text{ kg}$$
Total mass of the balloon filled with helium $= (100 + 1117.5)\text{ kg} = 1217.5\text{ kg}$
Payload $= (5028.6 – 1217.5)\text{ kg} = 3811.1\text{ kg}$
Hence, the payload of the balloon is $3811.1\text{ kg}$.

Question 17:

Calculate the volume occupied by $8.8\text{ g }\text{CO}_2$ at $31.1^\circ\text{C}$ and $1\text{ bar}$. Use $R = 0.083\text{ bar L K}^{-1}\text{ mol}^{-1}$.

Solution:

It is known that $pV = nRT = \frac{w}{M} RT$
Here, $m = 8.8\text{ g}$, $R = 0.083\text{ bar L K}^{-1}\text{ mol}^{-1}$, $T = 31.1^\circ\text{C} = 304.1\text{ K}$, $M = 44\text{ g}$, $p = 1\text{ bar}$
$$V = \frac{wRT}{pM} = \frac{8.8 \times 0.083 \times 304.1}{1 \times 44} = 5.048\text{ L} \approx 5.05\text{ L}$$
Hence, the volume occupied is $5.05\text{ L}$.

Question 18:

A $2.9\text{ g}$ sample of a gas at $95^\circ\text{C}$ occupies the same volume as $0.184\text{ g }\text{H}_2$ at $17^\circ\text{C}$ at the same pressure. What is molar mass of the gas.

Solution:

Volume ($V$) occupied by dihydrogen is given by:
$$V = \frac{n_{\text{H}_2} R T_{\text{H}_2}}{p} = \frac{(0.184 / 2) R (273 + 17)}{p} = \frac{0.092 \times 290 R}{p}$$
Let $M$ be the molar mass of the unknown gas. Volume ($V$) occupied by the unknown gas can be calculated as:
$$V = \frac{n_{\text{gas}} R T_{\text{gas}}}{p} = \frac{(2.9 / M) R (273 + 95)}{p} = \frac{2.9 \times 368 R}{M p}$$
According to the question, volumes and pressures are identical:
$$\frac{0.092 \times 290}{1} = \frac{2.9 \times 368}{M} \implies 26.68 = \frac{1067.2}{M} \implies M = \frac{1067.2}{26.68} = 40\text{ g mol}^{-1}$$
Hence, the molar mass of the gas is $40\text{ g mol}^{-1}$.

Question 19:

A mixture of dihydrogen and dioxygen at one bar pressure contains $20\%$ by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Solution:

Let the weight of dihydrogen be $20\text{ g}$ and the weight of dioxygen be $80\text{ g}$.
Then, the number of moles of dihydrogen:
$$n_{\text{H}_2} = \frac{20\text{ g}}{2\text{ g/mol}} = 10\text{ mol}$$
and the number of moles of dioxygen:
$$n_{\text{O}_2} = \frac{80\text{ g}}{32\text{ g/mol}} = 2.5\text{ mol}$$
Total moles $= 10 + 2.5 = 12.5\text{ mol}$.
Given, Total pressure of the mixture, $p_{\text{total}} = 1\text{ bar}$.
Then, partial pressure of dihydrogen:
$$p_{\text{H}_2} = x_{\text{H}_2} \times p_{\text{total}} = \left(\frac{10}{12.5}\right) \times 1\text{ bar} = 0.8\text{ bar}$$
Hence, the partial pressure of dihydrogen is $0.8\text{ bar}$.

Question 20:

What would be the SI unit for the quantity $\frac{pV^2 T^2}{n}$?

Solution:

The SI unit for pressure, $p$ is $\text{N m}^{-2}$.
The SI unit for volume, $V$ is $\text{m}^3$.
The SI unit for temperature, $T$ is $\text{K}$.
The SI unit for the number of moles, $n$ is $\text{mol}$.
Therefore, the SI unit for quantity is given by:
$$\frac{(\text{N m}^{-2}) (\text{m}^3)^2 (\text{K})^2}{\text{mol}} = \text{N m}^4\text{ K}^2\text{ mol}^{-1} \text{ (or } \text{kg m}^5\text{ s}^{-2}\text{ K}^2\text{ mol}^{-1}\text{)}$$

Question 21:

Using Charles’ law, explain why $-273.15^\circ\text{C}$ is the lowest possible temperature.

Solution:

Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature ($V \propto T$).

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in $^\circ\text{C}$) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at $-273.15^\circ\text{C}$. In other words, the volume of any gas at $-273.15^\circ\text{C}$ is zero theoretically. This is because all gases get liquefied before reaching a temperature of $-273.15^\circ\text{C}$. Hence, it can be concluded that $-273.15^\circ\text{C}$ is the lowest possible temperature.

Question 22:

Critical temperature for carbon dioxide and methane are $31.1^\circ\text{C}$ and $-81.9^\circ\text{C}$ respectively. Which of these has stronger intermolecular forces and why?

Solution:

Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature.
Hence, intermolecular forces of attraction are stronger in the case of $\text{CO}_2$ ($31.1^\circ\text{C} > -81.9^\circ\text{C}$).

Question 23:

Explain the physical significance of the van der Waals parameters.

Solution:

Physical significance of ‘a’:
‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.

Physical significance of ‘b’:
‘b’ is a measure of the volume of a gas molecule (excluded volume).