solution of 11 class chapter 5 states of matter
Answers
NCERT TEXTBOOK QUESTIONS SOLVED
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: 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. 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: 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 the gas pressure P.
Answer: According to ideal gas equation
PV = nRT or PV=nRT/V
ncert-solutions-for-class-11th-chemistry-chapter-5-states-of-matter-1
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: Using the 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 x 5(Molecular mass of N2 = 28 u)
or M1 = 70u
Question 5. Pressure of l 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: Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be
ncert-solutions-for-class-11th-chemistry-chapter-5-states-of-matter-2
Question 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 0.15g of aluminium reacts?
Answer: The chemical equation for the reaction is
2 Al + 2 NaOH + H20 -> 2 NaAl02 + 3H2 (3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H2 gas = 3 x 22400 0.15 g of Al at N.T.P release (the below pic is for 6th question continuation)
Question 8. What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?
Answer: Calculation of partial pressure of H2 in 1L vessel P1= 0.8 bar,
P2= ? V1= 0.5 L , V2 = 1.0 L
As temperature remains constant, P1V1 = P2V2
(0.8 bar) (0.5 L) = P2 (1.0 L) or P2 = 0.40 bar, i.e., PH2 = 0.40 bar
Calculation of partial pressure of 02 in 1 L vessel
P1‘ V1 = P2‘V2‘
(0.7 bar) (2.0 L) = P2 (1L) or P2‘ = 1.4 bar, i.e.,Po2= 1.4 bar
Total pressure =PHz + PQ2 = 0.4 bar + 1.4 bar = 1.8 bar
Question 23. Explain the physical significance of vander Waals parameters.
Answer: ‘a’ is a pleasure of the magnitude of the intermolecular forces of attraction, while b is a measure of the effective size of the gas molecules.
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