Question 391:
easyAt constant volume temperature is increased then: (1989)
An increase in temperature raises the thermal agitation (root-mean-square speed) of gas molecules. This causes them to travel faster, increasing the frequency of their collisions with the container walls.
Question 392:
easyThe volume occupied by the molecules contained in $4.5 \text{ kg}$ water at STP, if the intermolecular forces vanish away is: (2022)
At STP, 1 mole of an ideal gas occupies $22.4 \text{ L}$. The number of moles in $4.5 \text{ kg}$ of water is $n = \frac{4500 \text{ g}}{18 \text{ g/mol}} = 250 \text{ moles}$. The volume is $V = 250 \times 22.4 \text{ L} = 5600 \text{ L} = 5.6 \text{ m}^3$.
Question 393:
easyA cylinder contains hydrogen gas at pressure of $249 \text{ kPa}$ and temperature $27^\circ\text{C}$. Its density is : ($R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$) (2020)
Using the ideal gas law in terms of density, $P = \frac{\rho RT}{M}$, we get $\rho = \frac{PM}{RT}$. For hydrogen, $M = 2 \times 10^{-3} \text{ kg/mol}$. Thus, $\rho = \frac{249 \times 10^3 \times 2 \times 10^{-3}}{8.3 \times 300} = 0.2 \text{ kg/m}^3$.
Question 394:
easyAn ideal gas equation can be written as $P = \frac{\rho RT}{M_0}$ where $\rho$ and $M_0$ are respectively, (2020-Covid)
In the equation $P = \frac{\rho RT}{M_0}$, $\rho$ represents the mass density (mass per unit volume) of the gas, and $M_0$ is the molar mass of the gas.
Question 395:
easyIncrease in temperature of a gas filled in a container would lead to: (2019)
According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to its absolute temperature ($E_k \propto T$). Therefore, an increase in temperature increases its kinetic energy.
Question 396:
easyA given sample of an ideal gas occupies a volume $V$ at a pressure $P$ and absolute temperature $T$. The mass of each molecule of the gas is $m$. Which of the following gives the density of the gas? (2016 – II)
From the ideal gas equation $PV = NkT$, the number density is $n = \frac{N}{V} = \frac{P}{kT}$. The mass density $\rho$ is mass per unit volume, so $\rho = m \times n = \frac{Pm}{kT}$.
Question 397:
easyTwo vessels separately contain two ideal gases $A$ and $B$ at the same temperature, the pressure of $A$ being twice that of $B$. Under such conditions, the density of $A$ is found to be $1.5$ times the density of $B$. The ratio of molecular weight of $A$ and $B$ is: (2015 Re)
We know $M = \frac{\rho RT}{P}$. Given $T_A = T_B$, $P_A = 2P_B$, and $\rho_A = 1.5\rho_B$. The ratio of molecular weights is $\frac{M_A}{M_B} = (\frac{\rho_A}{\rho_B}) \times (\frac{P_B}{P_A}) = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4}$.
Question 398:
easyAt $10^\circ\text{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^\circ\text{C}$ this ratio is (2008)
Since $\frac{\rho}{P} = \frac{M}{RT}$, the ratio is inversely proportional to the absolute temperature $T$. Thus, $x_2 = x_1 (\frac{T_1}{T_2}) = x \times \frac{10 + 273}{110 + 273} = x \times \frac{283}{383}$.
Question 399:
easyThe equation of state for $5 \text{ g}$ of oxygen at a pressure $P$ and temperature $T$, when occupying a volume $V$, will be: (2004)
The number of moles $n = \frac{\text{given mass}}{\text{molar mass}}$. For $5 \text{ g}$ of $O_2$ gas, $n = \frac{5}{32}$. Substituting this into the ideal gas equation $PV = nRT$, we get $PV = (\frac{5}{32})RT$.
Question 400:
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.