**NCERT Solutions Class 11 Chemistry Chapter 5 States of Matter**

**1. What will be the minimum pressure required to compress 500 dm**^{3} of air at 1 bar to 200 dm^{3} at 30°C?

**Answer:**

**Given, **

Initial pressure, p_{1} = 1 bar

Initial volume, V_{1} = 500 dm^{3}

Final volume, V_{2} = 200 dm^{3}

Since the temperature remains constant, the final pressure (p_{2}) can be calculated using Boyle’s law.

**According to Boyle’s law,**

**Therefore, the minimum pressure required is 2.5 bar.**

**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, **

Initial pressure, p_{1} = 1.2 bar

Initial volume, V_{1} = 120 mL

Final volume, V_{2} = 180 mL

Since the temperature remains constant, the final pressure (p_{2}) can be calculated using Boyle’s law.

**According to Boyle’s law,**

**Therefore, the pressure would be 0.8 bar.**

**3. Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressurep.**

**Answer:**

The equation of state is given by,

pV = nRT ……….. **(i) **

Where,

p → Pressure of gas

V → Volume of gas

n→ Number of moles of gas

R → Gas constant

T → Temperature of gas

**From equation (i) we have,**

\(\frac{n}{V} = \frac p {RT}\)

Replacing n with \(\frac mM\) , we have

\(\frac mM=\frac p {RT}\) ............. **(ii)**

Where, m → Mass of gas

M → Molar mass of gas

But, \(\frac m V =d\)(d = density of gas)

**Thus, from equation (ii), we have**

\(\frac d M = \frac p {RT}\)

⇒ d = \((\frac M {RT})p\)

Molar mass (M) of a gas is always constant and therefore, at constant temperature

(T),\(\frac M {RT}\) = constant.

d = (constant) p

⇒ d \(\propto\) p

**Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p)**

**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:**

Density (d) of the substance at temperature (T) can be given by the expression,

d = \(\frac{Mp}{RT}\)

Now, density of oxide (d_{1}) is given by,

d_{1} = \(\frac{M_1p_1}{RT}\)

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} = \(\frac{M_2 p_2}{RT}\)

Where, M_{2} and p_{2} are the mass and pressure of the oxide respectively.

According to the given question,

d_{1} = d_{2}

\(\therefore\) M_{1}p_{1 }= M_{2}p_{2}

**GIven,**

p_{1} = 2 bar

p_{2} = 5 bar

Molecular mass of nitrogen, M_{2} = 28 g/mol

Hence, the molecular mass of the oxide is 70 g/mol.

**5. Pressure of 1 g 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 a relationship between their molecular masses.**

**Answer:**

For ideal gas A, the ideal gas equation is given by,

p_{A}V = n_{A}RT ..........**(i)**

Where, p_{A} and n_{A} represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

p_{B}V = n_{B}RT ..........**(ii)**

Where, p_{B} and n_{B} represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

**From equation (i), we have**

**From equation (ii), we have**

Where, M_{A} and M_{B} are the molecular masses of gases A and B respectively.

**Now, from equations (iii) and (iv), we have**

**Given,**

m_{A} = 1 g

p_{A} = 2 bar

m_{B} = 2 g

p_{B} = (3 - 2) = 1 bar

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

**Thus, a relationship between the molecular masses of A and B is given by 4M**_{A} = M_{B} .

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

**Answer:**

The reaction of aluminium with caustic soda can be represented as:

At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 mL of H_{2}.

\(\therefore\) 0.15 g Al gives \(\frac{3\times22400\times0.15}{54}\) mL of H_{2 }

i.e., 186.67 mL of H_{2}.

**At STP,**

p_{1 }= 1 atm

V_{1} = 186.67 mL

T_{1} = 273.15 K

Let the volume of dihydrogen be V_{2} at p_{2} = 0.987 atm (since 1 bar = 0.987 atm) and T_{2} = 20°C = (273.15 + 20) K = 293.15 K.

**Now,**

**Therefore, 203 mL of dihydrogen will be released.**