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NEET Chemistry MCQ
Gaseous State Mcq Neet Chemistry
Quiz 5
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Q.1
The temperature of an ideal gas is reduced from 927°C to 27°C. The r.m.s. velocity of the molecules becomes... [ Kerala CEE 2001]
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a) double the initial value
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b) half of the initial value
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c) four times the initial value
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d) ten times the initial value
Explanation
r.m. velocity ∝ √T , for same gas Answer: (b)
Q.2
the volume of an ideal gas is 44.8 lit at STP. Then at 267°C, What is its pressure is ...[ AFMC1998]
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a) 1 atm
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b) 2 atm
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c) 4 atm
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d) 6 atm
Explanation
STP condition 2 mole ideal gas will occupy 44.8 lit volume hence n=2PV=nRT P=(nRT) / V P=(2×8.314×540) / 44.8 P=200.42 kPa=1.987 atm Answer: (b)
Q.3
in gas equation PV=nRT the correct statement is ...[ AFMC 1997]
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a) P is the pressure of one mole of gas
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b) V is the volume of one mole of gas
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c) n is the number of moles of gas
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d) n is the number of molecules of gas
Explanation
Answer: (c)
Q.4
A gas can be liqified.. [ AFMC 2005]
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a) above its critical temperature
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b) at its critical temperature
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c) below its critical temperature
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d) at any temperature
Explanation
Critical temperature: The temperature above which a gas can not be liquefied at any pressure is called critical temperature. The corresponding pressure and volume is called critical pressure and critical volume respectively Answer: (c)
Q.5
At 27°C , the ratio of rms velocity of ozone to oxygen is ...(EAMCET 1992]
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a) √(3/5)
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b) √(4/3)
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c) √(2/3)
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d) 0.25
Explanation
for different gases at same temperature rms velocity ∝ 1/√M Answer: (c)
Q.6
If the average velocity of N2 molecules is 0.3 m/s at 27°C, then the velocity of 0.6 m/s will takes place at ..[ Manipal PMT 2001]
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a) 273 K
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b) 927 K
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c) 1000 K
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d) 1200 K
Explanation
r.m. velocity ∝ √T , for same gas Thus velocity doubles temperature becomes four times Answer: (d)
Q.7
Container A and B have same gases. pressure, volume and temperature of A are all twice that of B, then the ratio of number of molecules of And B are:
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a) 1 : 2
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b) 2 : 1
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c) 1 : 4
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d) 4 : 1
Explanation
for gas A:PA is pressure, VB is volume , TA is temperature then according gas law PAVA=nA R TASimilarly for gas B: PBVB=nB R TBDividing equation (1) by equation(2) we getNow 2PB=PA2VB=VA2TB=TASubtitling above values in equation we get nA=2nB Answer: (a)
Q.8
Diffusion of helium gas is
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a) Four times faster than ClO2
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b) Four times faster than NO2
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c) Four times faster than SO2
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d) Four times faster than CO2
Explanation
All options have rate four times Rate of diffusion of different gases are inversely proportional to square root of their densities ( molecular mass) Molecular mass of SO2 is 64 Answer: (c)
Q.9
What is the pressure of 2 moles of NH3 at 27°C when its volume is 5L in van der Waals equation? ( a=4.17, b=0.03711) [ Orissa JEE 2004]
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a) 10.33 atm
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b) 9.333 atm
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c) 9.74 atm
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d) 9.2 atm
Explanation
given a=4.17, n=2 moles, V=5L, b=0.03711R=0.0821 L atm K-1T=27°=300Kvan der Walls equation is Answer: (c)
Q.10
The rms velocity of an ideal gas at 27°C is 0.3 m/s. Its rms velocity at 927°C ( in m-1) is ..[ IIT 1986]
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a) 3.0
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b) 2.4
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c) 0.9
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d) 0.6
Explanation
For the same gas at different temperature rms velocity ∝ √T Answer: (d)
Q.11
The kinetic energy of 4 moles of nitrogen gas at 127°C is ... cals ( R=2 cal mol-1 K-1) [ EAMCET 2003]
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a) 4400
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b) 3200
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c) 4800
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d) 1524
Explanation
Answer: (c)
Q.12
Which one of the following statements is not true about the effect of an increase in temperature on the distribution of molecular speed in gas? [ AIEEE 2005]
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a) The most probable speed increases
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b) The fraction of the molecules with the most probable speed increases
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c) The distribution becomes broader
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d) The area under the distribution curve remains same as under the lower temperature
Explanation
According to Maxwll's distribution curve, as temperature increases, most probable velocity increases and fraction of molecules possessing most probable velocity decreases Answer: (b)
Q.13
At very high pressure, the compressibility factor of gas
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a) Increases with increase of temperature
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b) Decreases with increase of temperature
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c) Remains constant with change in temperature
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d) Is less than one
Explanation
At very high pressure, volume V is very small. a/V2 is large but in comparison to high P it can be neglected. therefore van der Waals equation becomes Z = PV/nRT. Thus, compressibility factor is greater than one and decreases with increase of temperature Answer: (b)
Q.14
When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1atm changes volume from 1 litre to 2 litre, the final temperature in Kelvin would be [ IIT 2005]
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a)
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b)
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c) T
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d)
Explanation
In case of monoatomic gas γ=5/3, thus γ-1=2/3So Answer: (a)
Q.15
The average kinetic energy of an ideal gas per molecule in SI units at 25°C will be ... [ CBSE PMT 1996]
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a) 6.17 × 10-21 kJ
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b) 6.17 × 10-21 J
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c) 6.17 × 10-20 J
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d) 7.16× 10-20 J
Explanation
Answer: (b)
Q.16
If r.m.s speed of gaseous molecules is x cm/sec at a pressure of p atm, then the r.m.s. speed at pressure of 2p atm and constant temperature will be .... [ Orissa JEE 2003]
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a) x
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b) 2x
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c) 4x
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d) x/4
Explanation
PV=constant as temperature is constant , Hence r.m.s. velocity is constant even if pressure is doubled Answer: (a)
Q.17
Pressure deviation from ideal behaviour takes place because of .. [ IIT 2003]
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a) Molecular interaction between atoms and PV/nRT > 1
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b) Molecular interaction between atoms and PV/nRT < 1
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c) Finite size of atom of PV/nRT > 1
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d) Finite size of atom of PV/nRT < 1
Explanation
pressure deviation from ideal gas behavior takes place because of intermolecular interactions. If we consider only pressure correction, then vander Waals equation reduces to Answer: (b)
Q.18
At moderate pressure, compressibility factor of gas
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a) Increases with increase in temperature
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b) Decreases with increase in temperature
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c) Remains constant with change in temperature
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d) Is less than one
Explanation
At moderate pressure, volume decreases, so that, a/V2can not be neglected However, 'b' can be neglected in comparision to V. Thus van der Waals equation reduces to Therefore, compressibility factor is less than one at moderate pressure and increases with increase of temperature Answer: (a)
Q.19
At very low pressure, the compressibility factor of CO2 having constant value of molar volume
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a) Increases with increase of molar volume
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b) Decreases with increase in temperature
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c) Remains constant with change in temperature
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d) is one
Explanation
According to van der Waals equation At very low pressure , volume (V) is very large. Hence, a/V2 is so small that it can be neglected. Similarly, volume correction term 'b' can also be neglected in comparison to V. Thus, the above equation reduces to PV=RT. HEnce, at very low pressure a real gas behaves like ideal gas ∴ Z=PV/RT=1 Answer: (d)
Q.20
Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is [ AIEEE 2007]
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a) 1/2
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b) 2/3
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c) (1/3) ×(273/298)
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d) 1/3
Explanation
let m be the mass of methane and oxygen Number of moles of oxygen=m/32 Number of moles of methane=m/16 Mole fraction of oxygen Let total pressure be P and Po be partial pressure of Oxygen then Po=XoP Po/P=Xo=(1/3) Answer: (d)
Q.21
In an experiment during the analysis of a carbon compound, 145 cm3 of H2 was collected at 760 mmHg pressure and 27°C temperature. The mass of H2 is nearly ... [ MLNR 1987]
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a) 10 mg
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b) 12 mg
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c) 24 mg
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d) 6 mg
Explanation
calculate volume at STP at STP 22400 cc=1 mole , number of moles=132/22400and mass=molecular weight × number of moles=12 mgAnswer: (b)
Q.22
What is the ratio of mean speed of an O3 molecule to the rms speed of an O2 molecule at the same temperature
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a) (3π/7)1/2
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b) (16/9π)1/2
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c) (3π)1/2
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d) (4π/9)1/2
Explanation
From the formula for mean speed , mean speed of O3 r.m.s velocity of O2 Answer: (b)
Q.23
he compressibility of a gas is less than unit at STP, therefore [ IIT 2000]
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a) Vm > 22.4 litres
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b) Vm < 22.4 litres
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c)Vm=22.4 litres
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d)Vm=44.8 litres
Explanation
When Z < 1, gas is more compressible and Vm < 22.4 L Z=PV/nRT , if Z < 1 it means, PV < nRT at STP, for one mole of a gas 1 atm × V < 1 × 0.0821 ×273 V < 22.4 LAnswer: (b)
Q.24
he total kinetic energy of 0.6 mol of an ideal gas at 27°C is
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a) 1681 J
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b) 2245 J
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c)1122 J
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d)2806 J
Explanation
K.E.=(3/2) nRT T=27°C=300 K K.E.=(3/2) × 0.6 × 8.314 × 300 K.E.=2244.7 J Answer:(b)
Q.25
In van der Waals equation of state of the gas law,the constant 'b' is measure of [ AIEEE 2004]
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a) Intermolecular repulsions
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b) Intermolecular collisions per unit volume
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c) Volume occupied by the molecules
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d) intermolecular attraction
Explanation
Answer: (c)
Q.26
Ratio of CP and CV of a gas 'X' is is 1.The number of atoms of the gas 'X' present in 11.2 litres of it at N.T.P is [ CBSE PMT 1989]
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a) 6.02 ×1023
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b) 1.2×1024
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c) 3.01 ×101023
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d) 2.01 ×1023
Explanation
Ratio of CP and CV of a gas is 1.4 thus gas is diatomic 11.2L at NTP=0.5 moleAnswer: (a)
Q.27
The term that corrects for the attractive forces present in a real gas in the van der Waals equation is [ IIT 2009]
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a)nb
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b) an2 / V2
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c)- (an2 / V2)
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d)-nb
Explanation
The pressure correction term n2 / V2 corresponds to the attractive forces among the molecules ( in the van der Waals equation)Answer: (b)
Q.28
The critical temperature of water is higher than that of O2 because the H2O molecules have [ IIT 1997]
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a) Fewer electrons than O2
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b) Two covalent bonds
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c)V - shape
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d)Dipole moment
Explanation
More the dipole moment more is the critical temperatureAnswer: (d)
Q.29
Which of the following expression is correct for an ideal gas?
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a)
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b)
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c)
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d)
Explanation
We known that average kinetic energy Answer: (c)
Q.30
The density of air is is 0.00130 g/ml. The vapour density of the air will be ...[ DEC 2000]
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a) 0.00065
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b) 0.65
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c) 14.4816
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d) 14.56
Explanation
Molecular weight=2 × V.D 22.4 L=volume of one mole 22.4 × density=2 × V.D.∴ V.D=11.2 × densityV.D=11200 cc × 0.0013=14.56Answer: (d)
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