Three 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)
1. $P < (P_1 + P_2 + P_3)$
2. $P = \frac{P_1 + P_2 + P_3}{3}$
3. $P = P_1 + P_2 + P_3$
4. $P > (P_1 + P_2 + P_3)$
View Answer
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, $P = P_1 + P_2 + P_3$.
According to kinetic theory of gases, at absolute zero of temperature: (1990)
1. Water freezes
2. Liquid helium freezes
3. Molecular motion stops
4. Liquid hydrogen freezes
View Answer
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.
Two 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)
1. 1 : 2
2. 1 : 1
3. 2 : 1
4. 4 : 1
View Answer
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.
At 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)
1. $5.016 \times 10^4\text{ K}$
2. $8.360 \times 10^4\text{ K}$
3. $2.508 \times 10^4\text{ K}$
4. $1.254 \times 10^4\text{ K}$
View Answer
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 $T = 8.360 \times 10^4\text{ K}$.
The molecules of a given mass of a gas have r.m.s velocity of $200\text{ ms}^{-1}$ at $27^{\circ}\text{C}$ and $1.0 \times 10^5\text{ Nm}^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ}\text{C}$ and $0.05 \times 10^5\text{ Nm}^{-2}$, the r.m.s. velocity of its molecules in $\text{ms}^{-1}$ is: (2016 – I)
1. $100\sqrt{2}$
2. $\frac{400}{\sqrt{3}}$
3. $\frac{100\sqrt{2}}{3}$
4. $\frac{100}{3}$
View Answer
RMS velocity depends only on temperature: $v_{rms} \propto \sqrt{T}$. Let $v_1$ and $v_2$ be rms speeds at $T_1 = 27^{\circ}\text{C} = 300\text{ K}$ and $T_2 = 127^{\circ}\text{C} = 400\text{ K}$. $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}}$. So $v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}\text{ ms}^{-1}$.
$4.0 \text{ g}$ of a gas occupies $22.4 \text{ litres}$ at NTP. The specific heat capacity of the gas at constant volume is $5.0 \text{ JK}^{-1}\text{mol}^{-1}$. If the speed of sound in this gas at NTP is $952 \text{ ms}^{-1}$, then the heat capacity at constant pressure is (Take gas constant $R = 8.3 \text{ J/mol K}$): (2015 Re)
1. $8.5 \text{ J/K mol}$
2. $8.0 \text{ J/K mol}$
3. $7.5 \text{ J/K mol}$
4. $7.0 \text{ J/K mol}$
View Answer
The density $\rho = \frac{4 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3}$. Using $v = \sqrt{\frac{\gamma P}{\rho}}$, we get $952 = \sqrt{\frac{\gamma \times 1.013 \times 10^5}{4/22.4}}$, yielding $\gamma \approx 1.6$. Since $\gamma = \frac{C_p}{C_v}$, $C_p = 1.6 \times 5.0 = 8.0 \text{ J K}^{-1} \text{mol}^{-1}$.
The amount of heat energy required to raise the temperature of $1 \text{ g}$ of Helium at NTP, from $T_1 \text{ K}$ to $T_2 \text{ K}$ is: (2013)
1. $\frac{3}{4} N_A k_B \left(\frac{T_2}{T_1}\right)$
2. $\frac{3}{8} N_A k_B (T_2 - T_1)$
3. $\frac{3}{2} N_A k_B (T_2 - T_1)$
4. $\frac{3}{4} N_A k_B (T_2 - T_1)$
View Answer
Helium is a monoatomic gas ($C_v = \frac{3}{2}R$). The number of moles in $1 \text{ g}$ is $n = \frac{1}{4}$. The heat required is $Q = n C_v \Delta T = \frac{1}{4} \left(\frac{3}{2}R\right)(T_2 - T_1) = \frac{3}{8} N_A k_B (T_2 - T_1)$.
If $C_p$ and $C_v$ denote the specific heats (per unit mass) of an ideal gas of molecular weight $M$ then: (2010 Mains)
1. $C_p - C_v = R/M^2$
2. $C_p - C_v = R$
3. $C_p - C_v = MR$
4. $C_p - C_v = \frac{R}{M}$
View Answer
Mayer's relation for molar heat capacities is $C_P - C_V = R$. Since molar heat capacity is related to specific heat capacity by $C = Mc$, we have $M C_p - M C_v = R$, which gives $C_p - C_v = \frac{R}{M}$.