Kinetic Theory Online Objective Test | Kinetic Theory Online Objective Test

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q1.

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 $^{\circ}$ C). The value of Molar volume is

A. 22.4 liters

B. 24.4 liters

C. 20.4 liters

D. 23.7 liters

Answer : Option A
Explaination / Solution:

PV=nRT

V=nRTP=1×8.31×2731.01×105=0.02224m3=22.4lit

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q2. Figure shows plot of PV/T versus P for 1.00

\times

1 0^{- 3}

kg of oxygen gas at two different temperatures The dotted plot corresponds to

A. ‘super heated’ gas behavior

B. ‘real’ gas behavior

C. ‘super cooled’ gas behavior

D. ‘ideal’ gas behavior

Answer : Option D
Explaination / Solution:

for 1 mole of ideal gas, according to ideal gas equation

$\begin{aligned} P V = R T \\ \frac{P V}{T} = R = c o n s t . \end{aligned}$

hence graph must be with zero slope. So that dotted line show ‘ideal’ gas behavior and curved line shows deviation from ‘ideal’ gas behavior

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q3. Figure shows plot of PV/T versus P for 1.00

\times

1 0^{- 3}

kg of oxygen gas at two different temperatures. Comparing

T_{1}

and

T_{2}

A. T1 = T2

B. T1 < T2

C. T1 > T2

D. T1 ≥ T2

Answer : Option C
Explaination / Solution:

for 1 mole of ideal gas, according to ideal gas equation

$\begin{aligned} P V = R T \\ \frac{P V}{T} = R = c o n s t . \end{aligned}$

hence graph must be with zero slope. So that dotted line show ‘ideal’ gas behavior and curved line shows deviation from ‘ideal’ gas behavior

A real gas behave as ideal gas at high temperature. Temperature T1 is close to dotted line.

so that $T_{1}$ > $T_{2}$

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q4.

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapor and other constituents) in a room of capacity 25.0 $m^{3}$ at a temperature of 27 $^{\circ}$ C and 1 atm pressure.

A. 6.60

\times

1 0^{26}

B. 6.50

\times

1 0^{26}

C. 6.30

\times

1 0^{26}

D. 6.10

\times

1 0^{26}

Answer : Option D
Explaination / Solution:

PV=NkTN=PVkT=1.01×105×251.38×10−23×300=6.1×1026

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q5. Estimate the average thermal energy of a helium atom at room temperature (27

^{\circ}

A. 6.4

\times

1 0^{- 21}

B. 6.0

\times

1 0^{- 21}

C. 5.7

\times

1 0^{- 21}

D. 6.2×10−21J

Answer : Option D
Explaination / Solution:

$E = \frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 10^{- 23} \times 300 = 6.2 \times 10^{- 21} J$

k = boltzman constant

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q6. Estimate the average thermal energy of a helium atom at the temperature on the surface of the Sun (6000 K)

A. 1.35

\times

1 0^{- 19}

B. 1.28

\times

1 0^{- 19}

C. 1.20

\times

1 0^{- 19}

D. 1.24×10−19J

Answer : Option D
Explaination / Solution:

$E = \frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 10^{- 23} \times 600 = 1.24 \times 10^{- 19} J$

k = boltzman constant

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q7. Estimate the average thermal energy of a helium atom the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

A. 1.9

\times

1 0^{- 16}

B. 2.3

\times

1 0^{- 16}

C. 2.1

\times

1 0^{- 16}

D. 2.0

\times

1 0^{- 16}

Answer : Option C
Explaination / Solution:

$E = \frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 10^{- 23} \times 10 \times 10^{6} = 2.07 \times 10^{- 16} J$

k = boltzman constant

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). The number of molecules

A. is the greatest in the vessel with neon

B. is the same in all three vessels

C. is the greatest in the vessel with uranium hexafluoride

D. is the greatest in the vessel with chlorine

Answer : Option B
Explaination / Solution:

Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q9.

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). In which case is $v_{r m s}$ the largest?

v_{r m s}

is largest for chlorine.

v_{r m s}

is largest for uranium hexafluoride.

v_{r m s}

is largest for neon.

v_{r m s}

is the same in all vessels.

Answer : Option C
Explaination / Solution:

$\begin{aligned} v_{r m s} = \sqrt{\frac{3 R T}{M}} \\ v_{r m s} \propto \frac{1}{\sqrt{M}} \end{aligned}$

Neon has smaller molecular wt in comparision to chlorine and uranium hexafluoride. So that $v_{r m s}$ is largest for neon.

# Kinetic Theory # CBSE 11TH PHYSICS Prepare / Learn

Q10.

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 $^{\circ} C$ ? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

A. 2.26

\times

1 0^{3}

B. 2.77

\times

1 0^{3}

C. 2.52

\times

1 0^{3}

D. 2.63

\times

1 0^{3}

Answer : Option C
Explaination / Solution:

vAr=vHe3RT39.9−−−−−√=3R×2534−−−−−−−−√T=39.9×2534=2.52×103K

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