When the temperature of an ideal gas is increased from 27°C to 927°C, the kinetic energy will be :

same
eight times
four times
twice

Solution

K.E. ∝ T [K.E. of one molecule of monoatomic gas = ³⁄₂RT]

As temperature increases to

\(\frac{927 + 273}{27 + 273} = \frac{1200}{300} = 4 \; times\)

therefore, Kinetic energy will be increased to 4 times.

Ratio of molecular weights of A and B is ⁴⁄₂₅ then ratio of rates of diffusion will be :

5 : 1
5 : 2
25 : 3
25 : 4

Solution

The rms speed at NTP of a gas can be calculated from the expression:

\(\sqrt{3P/d}\)
\(\sqrt{3PV/M}\)
\(\sqrt{3RT/M}\)
All of these

Solution

The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is

1.086 : 1
1 : 1.086
2 : 1.086
1.086 : 2

Solution

When universal gas constant(R) is divided by Avogadro no. (N₀), then the value of R/N₀ is equivalent to

Rydberg's constant
Boltzmann's constant
Planck's constant
Van der waal's constant

Solution

^R⁄_N₀ is also known as Boltz mann’s constant.

As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms changes by a factor of which of the following ?

³¹³⁄₂₉₃
\(\sqrt{313/293}\)
¹⁄₂
2

Solution

According to the kinetic theory of gases, in an ideal gas,between two successive collisions a gas molecule travels

in a wavy path
in a straight line path
with an accelerated velocity
in a circular path

Solution

According to kinetic theory the gas molecules are in a state of constant rapid motion in all possible directions colloiding in a random manner with one another and with the walls of the container and between two successive collisions molecules travel in a straight line path but show haphazard motion due to collisions.

Root mean square velocity of a molecule is 1000 m/s. The average velocity of the molecule is :

455.55 m/s
675.55 m/s
921.58 m/s
1221.58 m/s

Solution

The inversion temperature for van der Waal’s gas is :

\(T_{i} = \frac{a}{\left ( Rb \right )}\)
\(T_{i} = \left (\frac{2a}{Rb} \right )\)
\(T_{i} = 0.5T\)
\(T_{i} = \frac{a}{2\left ( R/b \right )}\)

Solution

The inversion temp is the temp below which the gas warms up on expansion

for, Vander Waal’s gas, \(T_{i} = \left (\frac{2a}{Rb} \right )\)

Which of the following expressions correctly represents the relationship between the average molar kinetic energy, \(\overline{K\!E}\), of CO and N₂ molecules at the same temperature ?

\(\overline{KE}_{co}\) < \(\overline{KE}_{N_{2}}\)
\(\overline{KE}_{co}\) > \(\overline{KE}_{N_{2}}\)
\(\overline{KE}_{co}\) = \(\overline{KE}_{N_{2}}\)
cannot be predicted unless volumes of the gases are given

Solution

Average molar kinetic energy = ³⁄₂kT

As temperature is same hence average kinetic energy of CO and N₂ is same.

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