<\/span><\/h3>\nAnswer:\u00a0<\/strong>P1<\/sub>\u00a0= 1 bar,P2<\/sub>\u00a0= ? V1<\/sub>= 500 dm3<\/sup>\u00a0,V2<\/sub>=200 dm3<\/sup>
As temperature remains constant at 30\u00b0C,
P1<\/sub>V1<\/sub>=P2<\/sub>V2<\/sub>
1 bar x 500 dm3<\/sup>\u00a0= P2<\/sub>\u00a0x 200 dm3<\/sup>\u00a0or P2<\/sub>=500\/200 bar=2.5 bar<\/p>\n<\/span>Question 2. A vessel of 120 mL capacity contains a certain amount of gas at 35\u00b0C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35\u00b0C. What would be its pressure?<\/strong><\/span><\/h3>\nAnswer: <\/strong>V1<\/sub>= 120 mL, P1<\/sub>=1.2 bar,
V2<\/sub>\u00a0= 180 mL, P2<\/sub>\u00a0= ?
As temperature remains constant, P1<\/sub>V1<\/sub>\u00a0= P2<\/sub>V2<\/sub>
(1.2 bar) (120 mL) = P2 (180mL)<\/p>\n<\/span>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.<\/strong><\/span><\/h3>\nAnswer:<\/strong>\u00a0According to ideal gas equation
PV = nRT or PV=nRT\/V
![\"a\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/07\/a.png\")
<\/p>\n<\/span>Question 4. At 0\u00b0C, 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?<\/strong><\/span><\/h3>\nAnswer:<\/strong>\u00a0Using the expression, d =MP\/RT , at the same temperature and for same density,
M1<\/sub>P1<\/sub>\u00a0= M2<\/sub>P2<\/sub>\u00a0(as R is constant)
(Gaseous oxide) (N2<\/sub>)
or
M1<\/sub>\u00a0x 2 = 28 x 5(Molecular mass of N2<\/sub>\u00a0= 28 u)
or M1<\/sub>\u00a0= 70u<\/p>\n<\/span>Question 5. Pressure of l g of an ideal gas A at 27\u00b0C 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.<\/strong><\/span><\/h3>\nAnswer:<\/strong>\u00a0Suppose molecular masses of A and B are MA<\/sub>\u00a0and MB<\/sub>\u00a0respectively. Then their number of moles will be
![\"b\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/07\/b.png\")
<\/p>\nQuestion 6. The drain cleaner, Drained contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 \u00b0C and one bar will be released when 0.15g of aluminums reacts?<\/b>
Answer:<\/strong>\u00a0The chemical equation for the reaction is
2 Al + 2 NaOH + H2<\/sub>0 -> 2 NaAl02<\/sub>\u00a0+ 3H2<\/sub>\u00a0(3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H2<\/sub>\u00a0gas = 3 x 22400 0.15 g of Al at N.T.P release
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q6.png\")
<\/p>\n![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q6.1.png\")
<\/p>\n
<\/span>Question 7. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm3<\/sup>\u00a0flask at 27 \u00b0C?<\/strong><\/span><\/h3>\nAnswer:<\/strong>
![\"c\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/07\/c.png\")
<\/p>\n<\/span>Question 8. What will be the pressure of the gas mixture when 0.5 L of H
![\"NCERT](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/07\/NCERT-Solutions-For-Class-11-Chemistry-Chapter-5-States-of-Matter.jpg\")
<\/p>\n