Without change in temperature, a gas is forced in a smaller volume. Its pressure increases because its molecules:

1.	strike the unit area of the container wall more often.
2.	strike the unit area of the container wall at a higher speed.
3.	strike the unit area of the container wall with greater force.
4.	have more energy.

A gas at pressure ${\mathrm{P}}_0$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be:

1. $4{\mathrm{P}}_0$

2. $2{\mathrm{P}}_0$

3. ${\mathrm{P}}_0$

4. $\frac{{\mathrm{P}}_0}{2}$

The mean free path for a gas, with molecular diameter \(d\) and number density \(n,\) can be expressed as:
1. \( \dfrac{1}{\sqrt{2} n \pi {d}^2} \)

2. \( \dfrac{1}{\sqrt{2} n^2 \pi {d}^2} \)

3. \(\dfrac{1}{\sqrt{2} n^2 \pi^2 d^2} \)

4. \( \dfrac{1}{\sqrt{2} n \pi {d}}\)

If at a pressure of \(10^6\) dyne/cm², one gram of nitrogen occupies \(2\times10^4\) c.c. volume, then the average energy of a nitrogen molecule in erg is:

What is the graph between volume and temperature in Charle's law?
1. An ellipse
2. A circle
3. A straight line
4. A parabola

If the pressure of a gas is doubled, then the average kinetic energy per unit volume of the gas will be:

The translational kinetic energy of \(n\) moles of a diatomic gas at absolute temperature \(T\) is given by:
1. \(\frac{5}{2}nRT\)
2. \(\frac{3}{2}nRT\)
3. \(5nRT\)
4. \(\frac{7}{2}nRT\)

Two isotherms are drawn at temperatures ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ as shown. The ratio of mean speed  at ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ is:
  

The translational kinetic energy of oxygen molecules at room temperature is \(60~\text J.\) Their rotational kinetic energy will be?
1. \(40~\text J\)
2. \(60~\text J\)
3. \(50~\text J\)
4. \(20~\text J\)

When the gas in an open container is heated, the mean free path:
1. Increases
2. Decreases
3. Remains the same
4. Any of the above depending on the molar mass