MENU

Chapter 5 States of Matter Solutions

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 - 1 : -

P1 = 1 bar,P2 = ?      V1= 500 dm3 ,V2=200 dm3
As temperature remains constant at 30°C,
P1V1=P2V2
1 bar x 500 dm3 = P2 x 200 dm3 or P2=500/200 bar=2.5 bar

Question - 2 : - Avessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 barpressure. The gas is transferred to another vessel of volume 180 mL at 35°C.What would be its pressure?

Answer - 2 : -

V1= 120 mL, P1=1.2 bar,
V2 = 180 mL, P2 = ?
As temperature remains constant, P1V1 =P2V2
(1.2 bar) (120 mL) = P2 (180mL)

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 - 3 : - According to ideal gas equation
PV = nRT or PV=nRT/V

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

Answer - 4 : -

Usingthe expression, d =MP/RT , at the same temperature and for same density,
M1P1 = M2P2 (as R is constant)
(Gaseous oxide) (N2)
or
M1 x 2 = 28 x5(Molecular mass of N2 =28 u)
or M1 = 70u

Question - 5 : - Pressure of lg of an ideal gas A at 27°C is found tobe 2 bar. When 2g of another ideal gas B is introduced in the same flask atsame temperature, the pressure becomes 3 bar. Find the relationship betweentheir molecular masses.

Answer - 5 : - Let MA and MB be the molar masses of the two gases Aand B. According to available data :

Question - 6 : - The drain cleaner, ‘Drainex’ contains small bits ofaluminium which react with caustic soda to produce hydrogen. What volume ofhydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts ?

Answer - 6 : - Step I. Calculation of thevolume of hydrogen released under N. T.P. conditions.
The chemical equation for the reaction is :
2Al + 2NaOH + 2H2O →2NaAl02 + 3H2
2×27=54 g Sod. meta aluminate 3×22400 mL
54g of Al at N.T.P. release H2 gas= 3 × 22400 mL
0.15g of Al at N.T.P. release H2 gas= (3×22400mL)×(0.15g)(54g) =186.7mL.
Step II.Calculation of volume of hydrogen released at 20°C and 1 bar pressure.

Question - 7 : - What will be the pressure exerted (in pascal) by amixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27°C ?

Answer - 7 : -

Question - 8 : - What will be the pressure of the gaseous mixturewhen 0.5 L of H2 at0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?

Answer - 8 : - Step I. Calculation ofpartial pressure of H2 in 1 L vessel.
V1, = 0-5 L V2 = 1.0 L
P1 = 0.8 bar P2 = ?
According to Boyle’s Law, P1V1 = P2V2
P2 = V2 = (0.8bar)×(0.5L)(1.0L) =0.4bar
Step II. Calculationof partial pressure of O2 in1 L vessel.
V1 = 2.0 L V2 = 1.0L
P1 = 0.7 bar P2 = ?
According to Boyle’s Law, P1V1 = P2V2
P2 = V2 = (0.7bar)×(2.0L)(1.0L) =1.4bar
Step III. Calculation of total pressure of gaseous mixture.
P = P1 +P2 = (0.4 + 1.4 ) bar =1.8 bar

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

Answer - 9 : -

Question - 10 : - 34.05 mL of phosphorus vapours weigh 0-0625 g at546°C and 1.0 bar pressure. What is the molar mass of phosphorus ?

Answer - 10 : - According to ideal gas equation,
PV = n RT ; PV = M orM = PV
Accordingto available data :
Mass of phosphorus vapours (W) = 0.0625 g
Volume of vapours (V) = 34.05 mL = 34.05 × 103 L
Pressure of vapours (P) = 1.0 bar.
Gas constant (R) = 0.083 bar L K-1 mol-1
Temperature (T) = 546 + 273 = 819 K

Free - Previous Years Question Papers
Any questions? Ask us!
NodeJS Interview Questions Answers
ASP.Net Interview Questions Answers
Interview Questions Answers
×