Question 141:
moderateAt $0\text{ K}$ which of the following properties of a gas will be zero? (1996)
According to kinetic theory of gases, at absolute zero ($0\text{ K}$), all molecular motion stops. Thus, kinetic energy becomes zero.
Practice NEET Thermal Physics Questions
Question 142:
moderateThree containers of the same volume contain three different gases. The masses of the molecules are $m_1$, $m_2$ and $m_3$ and the number of molecules in their respective containers are $N_1$, $N_2$ and $N_3$. The gas pressure in the containers are $P_1$, $P_2$ and $P_3$ respectively. All the gases are now mixed and put in one of these containers. The pressure $P$ of the mixture will be: (1991)
According to Dalton's law of partial pressures, the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Thus, $P = P_1 + P_2 + P_3$.
Question 143:
moderateAccording to kinetic theory of gases, at absolute zero of temperature: (1990)
The kinetic energy of gas molecules is directly proportional to absolute temperature ($E = \frac{3}{2}kT$). At absolute zero ($T = 0\text{ K}$), the kinetic energy is zero, meaning molecular motion stops.
Question 144:
moderateTwo containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of: (1988)
Vapour pressure of a liquid depends only on the nature of the liquid and its temperature. It is independent of the amount of liquid or the volume of the container. Since both are at the same temperature, the vapour pressure will be the same, so the ratio is 1:1.
Question 145:
moderateAt what temperature will the 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 $k_B = 1.38 \times 10^{-23}\text{ J K}^{-1}$) (2018)
Escape velocity $v_e = 11.2\text{ km/s} = 11200\text{ m/s}$. RMS speed $v_{rms} = \sqrt{\frac{3k_BT}{m}}$. Equating them: $11200 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}$. Solving for $T$ gives $T = 8.360 \times 10^4\text{ K}$.
Question 146:
moderateA gas mixture consists of 2 moles of $O_2$ and 4 moles of $Ar$ at temperature T. Neglecting all vibrational modes, the total internal energy of the system is: (2017-Delhi)
$O_2$ is diatomic ($f=5$), $U_1 = n_1\frac{f_1}{2}RT = 2 \times \frac{5}{2}RT = 5RT$. Ar is monoatomic ($f=3$), $U_2 = n_2\frac{f_2}{2}RT = 4 \times \frac{3}{2}RT = 6RT$. Total energy $U = U_1 + U_2 = 5RT + 6RT = 11RT$.
Question 147:
moderateThe molar specific heat at constant pressure of an ideal gas is $\frac{7}{2}R$. The ratio of specific heat at constant pressure to that at constant volume is: (2006)
Given $C_p = \frac{7}{2}R$. Using $C_p - C_v = R$, we find $C_v = \frac{7}{2}R - R = \frac{5}{2}R$. The ratio of specific heats is $\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5}$.
Question 148:
moderateFor hydrogen gas $C_p – C_v = a$ and for oxygen gas $C_p – C_v = b$, so the relation between $a$ and $b$ is given by: (1991)
Here $C_p$ and $C_v$ are specific heats per unit mass. For a gas, $C_p - C_v = \frac{R}{M}$. For hydrogen ($M=2$), $a = \frac{R}{2}$. For oxygen ($M=32$), $b = \frac{R}{32}$. Therefore, $a = 16b$.
Question 149:
moderateFor a certain gas the ratio of specific heats is given to be $\gamma = 1.5$. For this gas: (1990)
Given $\gamma = \frac{C_p}{C_v} = 1.5 = \frac{3}{2}$. In work units, $C_p - C_v = \frac{R}{J}$. Substituting $C_v = \frac{2}{3}C_p$, we get $C_p - \frac{2}{3}C_p = \frac{R}{J}$, leading to $\frac{C_p}{3} = \frac{R}{J}$, or $C_p = \frac{3R}{J}$.
Question 150:
moderateThe mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as: (2020)
From the kinetic theory of gases, the mean free path $\lambda$ (the average distance a molecule travels between collisions) is derived as $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$.