Rajasthan Board RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Textbook Exercise Questions and Answers.

Question 5.1.

A balloon is filled with hydrogen at room temperature. it will burst, if pressure exceed 0.2 bar. If at 1 bar pressure, the gas occupies 2.27 L volume, upto what volume can the balloon be expanded?

Answer:

According to Boyle's law, P_{1}V_{1} = P_{2}V_{2}

P_{1} = 1 bar, V_{1} = 2.27 L

If P_{2} = 0.2 bar then V_{2} = \(\frac{P_1 V_1}{P_2}\)

If \(=\frac{1 \times 2.27}{0.2}\)

V_{2 }= 1135 L

Question 5.2.

A ship sailing in pacific ocean consists of a balloon which is filled with 2L air, where temperature is 23.4°C; Now, what will be the value of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C ?

Answer:

Given that

V_{1} = 2L

T_{2} = 26.1 + 273 = 299.1 K

T (234 + 273) K = 2964 K

From Charles' law,

= 2L x 1.0092018 L

Question 5.3

At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL?

Answer:

Given,

P_{1} = 760 mm of Hg

V_{1} = 600 mL

T_{1} = 25 + 273 = 298 K

P_{2} =?

V_{2} = 640 mL

T_{2} = 10 + 273 = 283 K

According to combined gas Law,

= 6766 mm Hg

Question 5.4.

A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If the pressure of the mixture of gases in the cylinder is 25 bar, what is the partial pressure of dioxygen and neon in the mixture ? (Atomic mass of Ne = 20u)

Answer:

Mass of dioxygen = 70.6 g

Molecular mass of dioxygen = 32 g mol^{-1}

Number of moles of dioxygen = \(\frac{70.6 \mathrm{~g}}{32 \mathrm{~g} \mathrm{~mol}^{-1}}\)

= 2.21 mol

Mass of neon = 167.5g

Molecular mass of neon= 20 g mol^{-1}

Number of moles of neon = \(\frac{167.5 \mathrm{~g}}{20 \mathrm{~g} \mathrm{~mol}^{-1}}\)

= 8375 mol

Mole fraction of dioxygen = \(\frac{n_1}{n_1+n_2}\)

\(=\frac{2.21}{2.21+8.375}\)

\(=\frac{2.21}{10.585}\)

= 0.21

Mole fraction of neon = \(\frac{n_2}{n_1+n_2}\)

\(\begin{aligned} & =\frac{8.375}{2.21+8.375} \\ & =\frac{8.375}{10.585}=0.79 \end{aligned}\)

∵ Partial pressure of gas = mole fraction x total pressure

∴ Partial pressure of dioxygen = 0.21 × (25 bar)

= 5.25 bar

Partial pressure of neon = 0.79 × (25 bar) = 19.75 bar

Question 5.1.

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?

Answer:

Given,

P_{1} = 1 bar

V_{1} = 500 dm^{3}

V_{2} = 200 dm^{3}

P_{2} =?

From Boyle's law,

P_{1}V_{1} = P_{2}V_{2}

(1 bar) x (500 dm^{3}) = P_{2 }× (200 dm3)

= 2.5 bar

Question 5.2.

A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?

Answer:

Given,

V_{1} = 120 mL

P_{1} = 12 bar

V_{2 }= 180 mL

P_{2} =?

From Boyle's Law,

P_{1}V_{1} = P_{2}V_{2}

(1.2 bar) x (120 mL) = P_{2} x (180 mL)

= 0.8 bar

Question 5.3.

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

Answer:

Given,

PV = nRT

This equation shows that density of a gas is proportional to gas pressure P (d x P).

Question 5.4.

At 0°C, the density of a certain oxide of a gas'at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide ?

Answer:

Given,

P_{1} = 2 bar

P_{1} = 5 bar

M_{1} =?

M_{2} =28 (Molecular mass of dinitrogen = 28)

∵ \(d=\frac{P M}{R T}\)

For oxide of gas \(d_1=\frac{P_1 M_1}{R T}\)

For dinitrogen \(d_2=\frac{P_2 M_2}{R T}\)

∵ Density of both gases are same.

= 70 g mol^{-1}

Question 5.5.

Pressure of 1g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.

Answer:

At 27°C,

Pressure of 1 g of an ideal gas A = 2 bar

Pressure of 2 g of another ideal gas B = 3 bar

∴Pressure due to gas B = 3 - 2 = 1 bar

From ideal gas equation,

From equation (1) and (2)

\(\frac{2}{1}=\frac{m_B}{2 m_A}\)

MB = 4m A

Question 5.6.

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?

Answer:

The reaction is written as,

2Al + 2NaOH + 2H_{2}O_{2} → NaAlO_{2 }+ 3H_{2}↑

(27 x 2 = 54) (224 L x 3 at STP = 67.2 L)

∵ 54 g of aluminium gives 67.2 L of H_{2} at STP

∴ 0.15 g of aluminium will give (V_{1}) = \(\frac{67.2 \times 0.15}{54}\)

= 0.187 L of H_{2}

Given,

P_{1} =1atm

P_{1} bar = 0.987 atm

T_{1 }= 273 K

T_{2 }= 273 + 20 = 293 K

V_{2} =?

= 0.2033 L

= 2033 mL

Question 5.7.

What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27°C ?

Answer:

Mass of methane = 3.2 g

Molecular mass of methane (CH_{4})

= 12 + 4 = 16

∴ Number of moles of CH_{4}

Mass of Carbon dioxide = 4.4g

Molecular mass of CO_{2} = 12 + (2 x 16) = 44

∴Number of moles of CO_{2}

Volume of flask (V) = 9 dm^{3} = 9 x 10 m^{3}

T = 27 + 273 = 300 K

For Methane PCH_{4 }= \(\frac{n R T}{V}\)

= 5.54 x 10^{4} Pa

For CO_{2}

= 2.77 x 10^{4} Pa

Total pressure exerted by mixture

= PCH_{4} + PCO_{2}

= 5.54 x 10 + 277 × 10^{4}

= (5.54 + 2.77) × 10^{4}

= &31× 10^{4} Pa

Question 5.8.

What will be the pressure of the gaseous mixture when 0.5 L of Hg at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?

Answer:

Given, For H_{2},

V_{1} = 0.5L

V_{2 }= 1L

P_{1} = 0.8 bar

P2 =?

From Boyle's law,

P_{1}V_{1} = P_{2}V_{2}

(0.8 bar) x (0.5 L) = P_{2} × (1 L)

= 0.4 bar

For O_{2},

V_{1} = 20L

V_{2} = 1L

P_{1} = 0.7 bar

P_{2} =?

From Boyle's law,

P_{1}V_{1} = P_{2}V_{2}

(0.7 bar) x (2.0 L) = \(P_2=\frac{(0.7 \mathrm{bar}) \times(2.0 \mathrm{~L})}{(1 \mathrm{~L})}\) = 1.4 bar

∴ Total pressure of gaseous mixture

= PH_{2} + PO_{2} = 0.4 + 14

= 18 bar

Question 5.9.

Density of a gas is found to be 5.46 g/dm3 at 27°C and 2 bar pressure. What will be its density at STP?

Answer:

Given,

∴ We know that,

d_{1} = 5.46 g dm^{-3}

T_{1} = 27 + 273 = 300 K

P_{1} = 2 bar

T_{2} = 273 K

P_{2} = 1 bar

d_{2} =?

= 3.0 gdm^{-3}

Question 5.10.

34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus ?

Answer:

R = 0.083 bar dm^{3} k^{-1} mol^{-1}

PV = nRT i.e.,n = \(\frac{P V}{R T}\)

\(=\frac{1.0 \mathrm{bar} \times\left(34.05 \times 10^{-3} \mathrm{dm}^3\right)}{0.083 \text { bar } \mathrm{dm}^3 \mathrm{k}^{-1} \times 819 \mathrm{k}}\)

= 5 × 10^{-4} mol.

∴ Mass of 1 mole = \(\frac{0.0625}{5 \times 10^{-4}}\)

g = 125g

∴ Molar mass = 125 g mol^{-1}.

Question 5.11.

A student forgot to add the reaction mixture to the round bottom flask at 27°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 to be 477°C. What fraction of air would have been expelled out?

Answer:

Given, T_{1} = 27 + 273 = 300 K

T_{2} = 477 + 273 = 750K

From Charles' law,

∴ Volume of air expelled out

= 25 V_{1} - V_{1}

= 1.5 V_{1}

∴ Fraction of air expelled out

\(=\frac{1.5 V_1}{2.5 V_1}=\frac{3}{5}\)

Question 5.12.

Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm^{3} K^{-1}mol^{-1}).

Answer:

Given

P = 332 bar

V = 5 dm^{3}

R = 0.083 bar dm^{3} K^{-1}mol^{-1}

n = 4.0

T = ?

From ideal gas equation,

PV = nRT

= 50 K

Question 5.13.

Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer:

Mass of dinitrogen (N) = 1.4g

Molar mass of dinitrogen (N_{2}) = 28

∴ Number of moles of dinitrogen

Total number of electrons present in 0.05 moles of dinitrogen = 0.05 x 6022 × 10^{23} x 14

= 4.2154 x 10^{23 }electrons

Question 5.14.

How much time would it take to distribute one Avogadro's number of wheat grains, if 1010 grains are distributed each second?

Answer:

∵10^{10} grains of wheat are distributed in 1 s.

∴ 6.022 × 10^{23} grains of wheat will be distributed in

= 190956 x 10^{6} years

Question 5.15.

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27° C. R = 0.083 bar dm^{3} K_{1} mol^{-1}.

Answer:

For dioxygen,

mass of dioxygen = 8g

molar mass of dioxygen = 32

∴ Number of moles of dioxygen

= 0.25

For dioxygen,

mass of dihydrogen = 4 g

molar mass of dihydrogen = 2

∴ Number of moles of dihydrogen

Total number of moles (n) = 2 + 0.25 = 2.25 moles

V = 1 dm3

T = 27+ 273 = 300 K

R = 0.083 bar dm^{3} K^{-1} mol^{-1}

From ideal gas equation,

PV = nRT

∴ \(P=\frac{n R T}{V}\)

\(=\frac{(2.25 \mathrm{moles}) \times\left(0.083 \mathrm{bar} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})}{\left(1 \mathrm{dm}^3\right)}\)

= 56.025 bar

Question 5.16.

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 12kg m and R = 0.083bar LK^{-1 }mol^{-1}).

Answer:

Given

Radius of balloon (r) = 10 m

P = 1.66 bar

T = 27 + 273 = 300 K

density of air (d) = 1.2 kg m

R = 0.083 bar LK^{-1} mol^{-1}

Volume of balloon \((V)=\frac{4}{3} \pi r^3\)

\(=\frac{4}{3} \times \frac{22}{7} \times(10)^3\)

= 4190.48 m

Mass of displaced air = density x volume

= 1.2 × 4190.48

= 5028,6 kg

From ideal gas equation,

PV = nRT

\(\begin{aligned} & n=\frac{P V}{R T} \\ = & \frac{(1.66 \mathrm{bar}) \times\left(4190.48 \mathrm{~m}^3\right)}{\left(0.083 \mathrm{bar} \mathrm{L}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})} \end{aligned}\)

= 279.36 x 10^{3} moles

Mass of Helium = 279.36 × 10^{3} x 4

= 1117.44 kg

Mass of filled balloon = 100 + 1117.44

= 1217.44 kg

Pay load = 50286 - 1217.44

= 381116 kg

Question 5.17.

Calculate the volume occupied by &8 g of CO_{2 }at 31.1°C and 1 bar pressure. R = 0.083 bar L K^{1} mol^{-1}.

Answer:

Given, mass of CO_{2} (W) = 88 g

T = 31.1 + 273 = 3041 K

p = 1 bar

Molar mass of CO_{2} (M) = 12 + (2 x 16) = 44

R = 0.083 bar LK^{-1} mol^{-1}

V = ?

Number of moles of CO_{2},

= 5.05 L

Question 5.18.

2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?

Answer:

For a gas,

Mass of gas = 29 g

Let molar mass of gas = M g mol^{-1}

T = 95 + 273 = 368 K

Number of moles of gas \(\frac{2.9}{M}\)

∵ PV = nRT

\(P V=\frac{2.9}{M} \times R \times(368 \mathrm{~K})\) ........... (i)

For dihydrogen,

Mass of dihydrogen = 0.184 g

Molar mass of dihydrogen = 2

T = 17 + 273 = 290 K

Number of moles of dihydrogen = \(\frac{0.184}{2}\)

∵ PV = nRT

\(P V=\frac{0.184}{2} \times R \times(290 \mathrm{~K})\) ...... (ii)

Fromn equation (i) and (ii),

= 40 g mol-1

Question 5.19.

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

Answer:

∵ Mass of dihydrogen in 20% of dioxygen.

So,

Mass of dihydrogen = 20 g

Molar mass of dihydrogen = 2

∴ Number of moles of dihydrogen

\( n_{\mathrm{H}_2}=\frac{20}{2}\)

= 10 moles.

Mass of dioxygen = 80 g

Molar mass of dioxygen = 32

∴ Number of moles of dioxygen

\(n_{\mathrm{O}_2}=\frac{80}{32}\)

= 2.5 moles

Partial pressure of dihydrogen

= 0.8 bar

Question 5.20.

What would be the SI unit for the quantity PV_{2}T/n? PV_{2}T_{2} Nm 2 x (m^{3})2 × (K)_{2} mol

Answer:

SI unit of

= Nm^{4} k_{2}mol^{-1}

Question 5.21.

In terms of Charles' law explain why - 273°C is the lowest possible temperature?

Answer:

From Charles' law,

Hence, -273° C is the lowest temperature because below this temperature, the volume will become negative. This is meaningless.

Question 5.22.

Critical temperature for carbon dioxide and methane are 311°C and -819°C respectively. Which of these has stronger intermolecular forces and why?

Answer:

Carbon dioxide has high critical temperature than methane it means that CO_{2 }can be liquified at higher temperature more easily. So, CO has stronger intermolecular forces than methane.

Question 5.23.

Explain the physical significance of van der waals parameters.

Answer:

From van der waals equation,

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

where 'd' and 'b' = van der Waals parameters These parameters depend upon nature of gas.

Significance of 'd':

It gives the magnitude of attractive forces between the gaseous molecules. Its unit is L^{2}mol^{-2}. Greater the value of 'a', greater will be the intermolecular force of attraction between molecules of gas.

Significance of 'b':

It is excluded volume per mol of a gas. Its unit is L^{1} mol^{-1}. The volume of 'b' is four times the actual volume of the gas molecules.

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