State of matter practice question

Question-1

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3  at 300 c?

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?

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.

Question-4

At 00 C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecule mass of the oxide?

Question-5

Pressure of 1g of an ideal gas A at 270 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

Question-6

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 200c and one bar will be released when 0.15 g of aluminium reacts?

Question-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 270C?

Question-8

What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bars and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel are 270C?

Question-9

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

Question-10

34.05 mL of phosphorous vapour weighs 0.00625 g at 546o C and 0.1 bar pressure.  What is the molar mass of phosphorous?

Answers

1  2.5 bar.    2  0.8 bar  

3  Solution:

PV = nRT (or) P (Pressure of gas) = 

We know = ρ (density)
 P = n. ρ RT
 Pressure of gas (P) α density of the gas (d).

4 Solution:PV = nRT
(or) P (Pressure of gas) = 
We know = ρ (density)
 P = n. ρ RT
 Pressure of gas (P) α density of the gas (d).

5 Solution:

Amount of ideal gas, A = WA = 1 g
Its pressure, PA = 2bar
Amount of another gas, B = WB = 2 g
Total pressure of the flask = 3 bar
 Partial Pressure of gas B in the gas mixture = 3-2 = 1 bar
PV = nRT
    = 
For gas A, PA = 
2 = ………(i)
For gas B,
PB = 
1 = --------(ii)
Divide equation (i) by (ii)

or = 2
4MA = MB
The molecular mass of Mb is four times of Ma.

6 Solution:
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

2 × 27 g of Al                 3 × 22.4 litre of H2
= 54 g

54g Al gives = 3 × 22.4 litre H2 at N.T.P.

0.15 g Al gives = = 0.186 litre at N.T.P.

= 

Normal Pressure P1 = 1 atm             

P2 = 1 bar = 0.987 atm

V1 = 0.186L                                    

V2 =?

T1 = 273K                                      

T2 = 273 + 20 = 293K

=

V2 =  = 0.202 litre = 202 ml.

7  Solution:

 = 3.2 g

Mol. Weight of CH4 = 16

Moles of CH4 = =0.2

= 4.4 g

Mol. Weight of CO2 = 44

Moles of CO2 = = 0.1 mol

Total moles present in mixture = 0.2 + 0.1 = 0.3 mol

Volume of flask = 9 dm3 = 9 L



Temperature = 273 + 27 = 300 K

Now for pressure of mixture = P × V = nRT

P × 9 = 0.3 × 0.0821 × 300

P = =0.821 atom

0.821 × 1.01 × 105 Pa = 8.29 × 104 Pa.

8 Solution:
Volume of H2, (V1) 0.5 l

Pressure of H2, (P1) = 0.8 bar

Volume of oxygen, (V2) = 2.0 l

Pressure of O2, (P2) = 0.7 bar

Total Volume of the flask (V1 + V2) = 1l

Total pressure of the gaseous mixture is P

P (V1 + V2) = P1V1 + P2V2

P × 1 = (0.8 × 0.5) + (0.7 × 2.0)

P × 1 = (0.4 + 1.4)

Total Pressure of the mixture of gases , P = 1.8/1 = 1.8 bar.


9 Solution:


d1 = 5.46 g/dm3                 d2 = ?

T1 = 273 + 27 = 300 K        T2 =  273 K

P1 = 2 bar

 = 

d2 =  = 3g/dm3.

10Solution:
P = 0.1 bar , V = 34.05 mL = 0.03405

R = 0.0821 litre-atm K-1 mol-1

T = 273 + 546 = 819 K

PV = nRT

n = =
   = 0.00005 mol-1

Weight of phosphorous vapour = 0.00625g

Molar mass of phosphorus =  = 125 g

Share this post