r.m. velocity ∝ √T , for same gas Answer: (b)

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)

Answer: (c)

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)

for different gases at same temperature rms velocity ∝ 1/√M Answer: (c)

r.m. velocity ∝ √T , for same gas Thus velocity doubles temperature becomes four times Answer: (d)

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)

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)

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)

For the same gas at different temperature rms velocity ∝ √T Answer: (d)

Answer: (c)

According to Maxwll's distribution curve, as temperature increases, most probable velocity increases and fraction of molecules possessing most probable velocity decreases Answer: (b)

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)

In case of monoatomic gas γ=5/3, thus γ-1=2/3So Answer: (a)

Answer: (b)

PV=constant as temperature is constant , Hence r.m.s. velocity is constant even if pressure is doubled Answer: (a)

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)

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)

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)

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)

calculate volume at STP at STP 22400 cc=1 mole , number of moles=132/22400and mass=molecular weight × number of moles=12 mgAnswer: (b)

From the formula for mean speed , mean speed of O3 r.m.s velocity of O2 Answer: (b)

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)

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)

Answer: (c)

Ratio of CP and CV of a gas is 1.4 thus gas is diatomic 11.2L at NTP=0.5 moleAnswer: (a)

The pressure correction term n2 / V2 corresponds to the attractive forces among the molecules ( in the van der Waals equation)Answer: (b)

More the dipole moment more is the critical temperatureAnswer: (d)

We known that average kinetic energy Answer: (c)

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)