The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

√2v
v
2v
4v

Solution

\(v_{rms}=\sqrt{\frac{3RT}{M}}\) According to problem T will become 2T and M will becomes M/2 so the value of vrms will increase by √4 = 2 times i.e. new root mean square velocity will be 2v.

The velocity of the molecules of a gas at temperature 120 K is v. At what temperature will the velocity be 2 v?

120 K
240 K
480 K
1120 K

Solution

At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is (T= absolute temperature of the gas) is

T²
T
1/T
1/T²

Solution

The adjoining figure shows graph of pressure and volume of a gas at two tempertures T₁ and T₂. Which of the following inferences is correct?

T₁ > T₂
T₁ = T₂
T₁ < T₂
None of these

Solution

For a given pressure, volume will be more if temperature is more (Charle's law)

From the graph it is clear that V₂ > V₁ ⇒ T₁ > T₂

The molecules of a given mass of gas have a root mean square velocity of 200m s^-1 at 27°C and 1.0 × 10⁵ N m^-2 pressure. When the temperature is 127°C and the pressure 0.5 × 10⁵ Nm^-2, the root mean square velocity in ms^-1, is

\(\frac{400}{\sqrt{3}}\)
100√2
\(\frac{100\sqrt{2}}{3}\)
\(\frac{100}{3}\)

Solution

If one mole of a monatomic gas(γ= 5/3) is mixed with one mole of a diatomic gas (γ= 7/3), the value of g for the mixture is

1.40
1.50
1.53
3.07

Solution

A fixed mass of gas at constant pressure occupies a volume V.The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be