Q: A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2 is :[Use atomic masses (in gmol−1) : N = 14, Ar = 40]

nN2 =287=41=0.25 nAr=408=51=0.20 Now, Applying Dalton’s law of partialpressure, pN2=(χN2)PTotal =0.450.25×27 bar =95×27=15 bar

Q: A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is :

• Compressibility factor(Z) =VidealVreal ∴Vreal<Videal; Hence Z • If Z < 1, attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

Q: Given van der Waals constant for NH3,H2,O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?

van der waal constant 'a', signifies intermolecular forces of attraction.- Higher is the value of ' a ', easier will be the liquefaction of gas.

Q: The correction factor ‘a’ to the ideal gas equation corresponds to :

In real gas equation, (P+V2an2)(V−nb)=nRT van der Waal's constant, 'a' signifies intermolecular forces of attraction.

Q: Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escape in the time required for one-half of the hydrogen to escape?

t2n2t1n1=M1M2⇒n2=21,n1=n′ ⇒t1x12t2n′=M1M2=322=161=41 Assuming t2=t1 t12n′t2=41 ∴n′=81

Q: Equal masses of H2,O2 and methane have been taken in a container of volume V at temeprature 27∘C in identical conditions. The ratio of the volumes of gases H2:O2: methane would be :

According to Avogadro's hypothesis volume ∝ molesnH2=2wnO2=32wnCH4=16w So, ratio is 2w:32w:16w =16:1:2

Q: Maximum deviation from ideal gas is expected from:

NH3 is polar molecule∴ intermolecular force of attraction are very high∴ a is maximum and deviation is maximum

Q: A 20 litre container at 400 K contains CO2( g) at pressure]0.4 atm and an excess of SrO (neglect the volume of solid SrO ). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be]?(Given that : SrCO3( s)⇌SrO(s)+CO2( g) .Kp=1.6 atm)

Max. pressure of CO2= Pressure of CO2 at equilibrium For reaction,SrCO3( s)⇌SrO(s)+CO2 Kp=PCO2=1.6 atm= maximum pressure of CO2 Volume of container at this stage,V=PnRT Since container is sealed and reaction was not earlier at equilibrium ∴n= constant n=RTPV=RT0.4×20 Put equation (ii) in equation (i)V=[RT0.4×20]1.6RT=5 L

Q: A gas such as carbon monoxide would be most likely to obey the ideal gas law at:

Real gases show ideal gas behaviour at high temperatures and low pressures.

Q: What will be the imum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 300 C?

We know, PV = P'V'By putting all those values, we get-The minimum pressure required to compress air is P′=V′PV= 2001×500=2.5bar.

Question 1

A mixture of N2​ and Ar gases in a cylinder contains 7 g of N2​ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2​ is :[Use atomic masses (in gmol−1) : N = 14, Ar = 40]

Accepted Answer

nN2​ =287​=41​=0.25 nAr​=408​=51​=0.20 Now, Applying Dalton’s law of partialpressure, pN2​​=(χN2​)PTotal​ 						=0.450.25​×27 bar						=95​×27=15 bar

Question 2

A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is :

Accepted Answer

• 	Compressibility factor(Z) =Videal​Vreal​​ ∴Vreal​<Videal​; Hence Z • 	If Z < 1, attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

Question 3

Given van der Waals constant for NH3​,H2​,O2​ and CO2​ are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?

Accepted Answer

van der waal constant 'a', signifies intermolecular forces of attraction.- Higher is the value of ' a  ', easier will be the liquefaction of gas.

Question 4

The correction factor ‘a’ to the ideal gas equation corresponds to :

Accepted Answer

In real gas equation, (P+V2an2​)(V−nb)=nRT  van der Waal's constant, 'a' signifies intermolecular forces of attraction.

Question 5

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escape in the time required for one-half of the hydrogen to escape?

Accepted Answer

t2​n2​​t1​n1​​​=M1​M2​​​⇒n2​=21​,n1​=n′ ⇒t1​x1​2t2​n′​=M1​M2​​​=322​​=161​​=41​ Assuming t2​=t1​ t1​2n′t2​​=41​ ∴n′=81​

Question 6

Equal masses of H2​,O2​  and methane have been taken in a container of volume V at temeprature 27∘C  in identical conditions. The ratio of the volumes of gases H2​:O2​:  methane would be :

Accepted Answer

According to Avogadro's hypothesis volume ∝  moles​nH2​​=2w​nO2​​=32w​nCH4​​=16w​​ So, ratio is 2w​:32w​:16w​  =16:1:2

Question 7

Maximum deviation from ideal gas is expected from:

Accepted Answer

NH3​  is polar molecule∴  intermolecular force of attraction are very high∴  a is maximum and deviation is maximum

Question 8

A 20 litre container at 400 K  contains CO2​( g) at pressure]0.4 atm  and an excess of SrO  (neglect the volume of solid SrO  ). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2​ attains its maximum value, will be]?(Given that : SrCO3​( s)⇌SrO(s)+CO2​( g) .Kp​=1.6 atm)

Accepted Answer

Max. pressure of CO2​= Pressure of CO2​  at equilibrium For reaction,SrCO3​( s)⇌SrO(s)+CO2​ Kp​=PCO2​​=1.6 atm= maximum pressure of CO2​ Volume of container at this stage,V=PnRT​ Since container is sealed and reaction was not earlier at equilibrium
∴n= constant  
n=RTPV​=RT0.4×20​
Put equation (ii) in equation (i)V=[RT0.4×20​]1.6RT​=5 L

Question 9

A gas such as carbon monoxide would be most likely to obey the ideal gas law at:

Accepted Answer

Real gases show ideal gas behaviour at high temperatures and low pressures.

Question 10

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

Accepted Answer

We know, PV = P'V'By putting all those values, we get-The minimum pressure required to compress air is P′=V′PV​=  2001×500​=2.5bar.

Question 11

A vessel of 120 mL capacity contain amount of gas at 35oC and 1.2 bar pressure The gas is transferred to another vessel of volume 180 mL at 35oC. What would be its pressure?

Accepted Answer

The temperature and the amount of gas is constant. Hence, pressure is calculated from Boyle's law.P′=V′PV​=  1801.2×120​=0.8bar

Question 12

Using the equation of state pV=nRT, show that, at a given temperature the density of a gas is proportional to its gas pressure p.

Accepted Answer

PV=nRT=Mw​RTHence,  PM=Vw​RT=ρRTρ∝PHere, ρ=Vw​n=Mw​ρ is the density, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, T is the temperature and M is the molar mass.

Question 13

At 0∘C the density of certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Accepted Answer

The expression for density is ρ=RTMP​.When the temperature is constant, M′=P′MP​.Here, M and P are the molar mass and pressures of nitrogen and M' and P' are the molar mass and pressures of oxide.Substitute values in the above equation.M′=228×5​=70 u

Question 14

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 a relationship between their molecular masses?

Accepted Answer

Let M and M' be the molar masses of ideal gases A and B respectively.The number of moles of gas A and B are  M1​ and M′2​ respectively.Let P and P' be the pressures of gases A and B respectively. P=2 bar P+P′=3 bar P′=3−2=1 barThe ideal gas equations for two gases A and B are,PV=nRT......(i)P′V=n′RT......(ii)Divide equation (i) by equation (ii). P′P​=n′n​=2×M1×M′​=2MM′​MM′​= 2P′P​ =  12×2​=4 M′=4M

Question 15

The drain cleaner, Drainex contains small bits of aluum 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 aluum reacts?

Accepted Answer

The balanced chemical equation is 2Al+2NaOH+2H2​O→2NaAlO2​+3H2​.54 g (2 mol) of Al will give 3 moles (or 67.2 L) of hydrogen.0.15 g of Al will give 5467.2​×0.15=0.189L of hydrogen.The volume of hydrogen is at 1 bar and 273 K. The volume at 1 bar and 200C will be, V′= P′TPVT′​=  1×2731×0.189×293​=0.2028L=202.8 ml.

Question 16

What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of CO2​ contained in a 9 dm3 flask at 27 C.

Accepted Answer

The number of moles of methane is 163.2​=0.2.The pressure exerted by methane is P=VnRT​=  9×10−30.2×8.314×300​=5.54×104Pa.The number of moles of CO2​ is 444.4​=0.1.The
 pressure exerted by CO2​ is P′=Vn′RT​=  9×10−30.1×8.314×300​=2.77×104Pa.The total pressure is P+P′=5.54×104Pa+2.77×104Pa=8.31×104Pa.

Question 17

What will be the pressure of the gaseous mixture when 0.5 L of H2​ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27 ∘C ?

Accepted Answer

(i) The partial pressure of hydrogen gas  P′=V′PV​=  10.8×0.5​=0.40 bar(ii) The partial pressure of oxygen gas  P′′=V′PV​=  10.7×2.0​=1.40 barSo, the total pressure of gaseous mixture is P+P′′=0.40+1.40=1.80 bar

Question 18

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

Accepted Answer

The expression for density is d=RTPM​.For a given gas, R and M are constant.d∝P/Td′=PT′dP′T​=  2×2735.46×1×300​=3.0 g dm−3

Question 19

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 ∘C and 0.1 bar pressure. What is the molar mass of phosphorus?

Accepted Answer

The molar mass of phosphorus, M=PVwRT​=  1×34.05×10−30.0625×0.08314×819​=124.98.

Question 20

A student forgot to add the reaction mixture to the round bottomed flask at 27oC but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477oC. What fraction of air would have been expelled out?

Accepted Answer

The initial temperature T=27oC=27+273K=300KThe final temperature  T′=477oC=477+273K=750KThe inital volume is V L and the final volume is V' L.V′V​=  T′T​V′V​=  750300​V′V​=  5′2​The fraction of air expelled is 1−V′V​=1−52​=  53​.

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Solution for Class 11, Chemistry, States of matter

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