At what temperature will the \(\text{rms}\) speed of oxygen molecules become just sufficient for escaping from the earth's atmosphere?
(Given: Mass of oxygen molecule \((m)= 2.76\times 10^{-26}~\text{kg}\), Boltzmann's constant \(k_B= 1.38\times10^{-23}~\text{J K}^{-1}\))
1. \(2.508\times 10^{4}~\text{K}\)
2. \(8.360\times 10^{4}~\text{K}\)
3. \(5.016\times 10^{4}~\text{K}\)
4. \(1.254\times 10^{4}~\text{K}\)
Two vessels separately contain two ideal gases \(\mathrm{A}\) and \(\mathrm{B}\) at the same temperature, the pressure of \(\mathrm{A}\) being twice that of \(\mathrm{B}\). Under such conditions, the density of \(\mathrm{A}\) is found to be \(1.5\) times the density of \(\mathrm{B}\). The ratio of molecular weight of \(\mathrm{A}\) and \(\mathrm{B}\) is:
1. | \(\frac{2}{3}\) | 2. | \(\frac{3}{4}\) |
3. | \(2\) | 4. | \(\frac{1}{2}\) |
One mole of an ideal diatomic gas undergoes a transition from \(A\) to \(B\) along a path \(AB\) as shown in the figure.
The change in internal energy of the gas during the transition is:
1. | \(20\) kJ | 2. | \(-20\) kJ |
3. | \(20\) J | 4. | \(-12\) kJ |
The ratio of the specific heats \(\frac{{C}_{{P}}}{{C}_{{V}}}=\gamma\) in terms of degrees of freedom(\(n\)) is given by:
1. | \(\left(1+\frac{1}{n}\right )\) | 2. | \(\left(1+\frac{n}{3}\right)\) |
3. | \(\left(1+\frac{2}{n}\right)\) | 4. | \(\left(1+\frac{n}{2}\right)\) |
The mean free path of molecules of a gas (radius \(r\)) is inversely proportional to:
1. \(r^3\)
2. \(r^2\)
3. \(r\)
4. \(\sqrt{r}\)
In the given \({(V\text{-}T)}\) diagram, what is the relation between pressure \({P_1}\) and \({P_2}\)?
1. | \(P_2>P_1\) | 2. | \(P_2<P_1\) |
3. | cannot be predicted | 4. | \(P_2=P_1\) |
The amount of heat energy required to raise the temperature of \(1\) g of Helium at NTP, from \({T_1}\) K to \({T_2}\) K is:
1. \(\frac{3}{2}N_ak_B(T_2-T_1)\)
2. \(\frac{3}{4}N_ak_B(T_2-T_1)\)
3. \(\frac{3}{4}N_ak_B\frac{T_2}{T_1}\)
4. \(\frac{3}{8}N_ak_B(T_2-T_1)\)
At \(10^{\circ}\mathrm{C}\) the value of the density of a fixed mass of an ideal gas divided by its pressure is \(x\). At \(110^{\circ}\mathrm{C}\) this ratio is:
1. | \(x\) | 2. | \(\dfrac{383}{283}x\) |
3. | \(\dfrac{10}{110}x\) | 4. | \(\dfrac{283}{383}x\) |
An increase in the temperature of a gas-filled in a container would lead to:
1. decrease in intermolecular distance.
2. increase in its mass.
3. increase in its kinetic energy.
4. decrease in its pressure.