Class 11 Chemistry

# States of Matter

## NCERT Solution

### Part 1

Question 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, p1 = 1 bar, V1 = 500 dm3, V2 = 200 dm3

As per Boyle’s Law: p1V1 = p2V2

Or, 1xx500=p_2xx200

Or, p_2=(500)/(200)=2.5 bar

Question 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, p1 = 1.2 bar, V1 = 120 mL, V2 = 180 mL

As per Boyle’s Law: p1V1 = p2V2

Or, 1.2xx120=p_2xx180

Or, p_2=(1.2xx120)/(180)=0.8 bar

Question 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: pV=nR\T

Or, p=(nR\T)/V

We know that n = constant mass of gas ÷ molar mass of gas =m/M

So, p=(mR\T)/(MV)

Since m/V=ρ

So, p=(ρR\T)/M

So, p∝ρ proved

Question 4: At 0° C, the density of a certain dioxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Answer: Molar mass of dinitrogen = 28 u

From previous question, we found

p=(ρR\T)/M

Or, ρ=(pM)/(RT)

As R is a constant and given temperature is same

So, p_1M_1=p_2M_2

Or, 2xx\M_1=5xx28

Or, M_1=(5xx28)/2=70 u

Question 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 the relationship between their molecular masses.

Answer: Let us assume that molar masses of A and B are MA and MB respectively. So, their number of moles can be as follows:

n_A=1/(M_A)

n_B=2/(M_B)

Given, pA = 2 bar, pA + pB = 3 bar

So, pB = 3 – 2 = 1 bar

Using the equation, pV = nRT, we get

p_AV=n_A\RT

And, p_BV=n_B\RT

So, (p_A)/(p_B)=(n_A)/(n_B)

=(1/(M_A))/(2/(M_B))

Or, (p_A)/(p_B)=(M_B)/(2M_A)

Or, (M_B)/(M_A)=(2p_A)/(p_B)=2xx2/1=4

Or, M_B=4M_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° C and one bar will be released when 0.15 g of aluminum reacts?

Answer: The chemical reaction of this process is given by following equation:

2Al + 2NaOH + H2 → 1NaAlO2 + 3H2

In this reaction, 2 M of aluminium produces 3 M of dihydrogen gas.

Molar mass of Al = 2 × 27 = 54 g

Volume of hydrogen produced at STP = 3 × 22720 mL

So, volume of hydrogen produce on reaction with 0.15 g

=(3xx22720xx0.15)/(54)=189.33 mL

Now, we have: V1 = 189.33 mL, p1 1 bar, p2 = 1 bar, T1 = 273 K and T2 = 20 + 273 = 293 K

Volume V2 can be calculated as follows:

According to ideal gas equation:

(p_1V_1)/(T_1)=(p_2V_2)/(T_2)

Or, V_2=(p_1V_1T_2)/(p_2T_1)

=(1xx189.33xx293)/(1xx273)=203 mL

Question 7: What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contianed in a 9 dm3 flask at 27° C?

Answer: Molar mass of methane = 16 u and molar mass of CO2 = 24 u

So, n of methane = (3.2)/(16)=0.2

And, n of carbon dioxide = (4.4)/(44)=0.1

R = 8.314 J K-1 mol-1

Given temperature = 27 + 273 = 300 K

Using pV = nRT

p=(nR\T)/V

So, pmethane =(0.2xx0.08314xx300)/9=0.55 atm

And, pcarbon dioxide =(0.1xx0.08314xx300)/9=0.27 atm

So, pressure exerted by both gases together = 0.55 + 0.27 = 0.82 atm

Question 8: What will the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of of dioxygen at 0.7 bar are introduced in a 1L vessel at 27° C?

Answer: Calculation of partial pressure of H2: p1 = 0.8 bar, p2 = ?, V1 = 0.5 L and V2 = 1 L

p_1V_1=p_2V_2

Or, 0.8xx0.5=p_2xx1

Or, p_2=0.8xx0.5=0.4 bar

Calculation of partial pressure of O2: p1 = 0.7 bar, p2 = ?, V1 = 2 L and V2 = 1 L

p_1V_1=p_2V_2

Or, 0.7xx2=p_2xx1

Or, p_2=1.4 bar

Total pressure = 0.4 + 1.4 = 1.8 bar

Question 9: Density of a gas is found to be 5.46 g/dm3 at 27° C at 2 bar pressure. What will be its density at STP?

Answer: We know: d=p/(RT)

So, (d_1)/(d_2)=(p_1T_2)/(p_2T_1)

Given, d1 = 5.46, p1 = 2 bar,T1 = 300 K, p2 = 1 bar,T2 = 273 K

We have, (5.46)/(d_2)=(2xx273)/(1xx300)

Or, d_2=(5.46xx300)/(2xx273)=3 g/dm3

Question 10: 34.05 mL of phosphorus vapor weighs 0.0625 g at 546° C and 0.1 bar pressure. What is the molar mass of phosphorus?

Answer: Given: V1 = 34.05 mL, T1 = 546 + 273 = 819 K, p1 = 0.1 bar, m = 0.0625 g

Let us calculate the volume at STP (T2 = 273 K, p2 = 1 bar

(p_1V_1)/(T_1)= (p_2V_2)/(T_2)

Or, (0.1xx34.05)/(819)=(1xx\V_2)/(273)

Or, V_2=(0.1xx34.05xx273)/(819)=1.135

Since 11.35 mL weighs 0.0625 g at STP

So, 22720 mL weighs (0.0625)/(11.35)xx22720=125 g

So, molar mass = 125 g mol-1