Question

A long cylindrical glass tube, equipped with a porous disc at the centre, contain methane gas at 5.0 atmosphere on one side and He gas at 2.0 atmosphere on the other side of the disc as shown in the diagram below:

Disc is permeable to both gases and rate of diffusion is directly proportional to the gas pressure and inversely proportional to square root of molar masses as:

$-\frac{dp}{dt} = k\frac{P}{\sqrt{M}}$ where, $k$ is a constant.

If $k$ for the diffusion of methane gas is $2.5 \times 10^{-2}$ second$^{-1}$, determine time after which pressure of methane chamber will drop to 4.0 atmospheres (nearest integer).

Answer 1

9

Answer 2

The rate of diffusion is given by $-\frac{dP}{dt} = k \frac{P}{\sqrt{M}}$. The problem states that "If $k$ for the diffusion of methane gas is $2.5 \times 10^{-2}$ second$^{-1}$". The units of the given value are s$^{-1}$. Let's analyze the units of the terms in the formula. The rate of change of pressure $\frac{dP}{dt}$ has units of pressure per unit time (e.g., atm/s). Pressure $P$ has units of pressure (atm). Molar mass $M$ has units of mass per mole (g/mol). Thus, $\sqrt{M}$ has units of $\sqrt{\text{g/mol}}$. The term $\frac{P}{\sqrt{M}}$ has units of $\frac{\text{atm}}{\sqrt{\text{g/mol}}}$. For the equation $-\frac{dP}{dt} = k \frac{P}{\sqrt{M}}$ to be dimensionally consistent, the constant $k$ must have units of $\frac{\text{atm/s}}{\text{atm/}\sqrt{\text{g/mol}}} = \frac{\sqrt{\text{g/mol}}}{\text{s}}$.

However, the given value $2.5 \times 10^{-2}$ has units of s$^{-1}$. This suggests that the given value is not the constant $k$ in the formula, but rather the effective rate constant for the diffusion of methane, which is proportional to $\frac{k}{\sqrt{M_{CH_4}}}$. Let's assume the rate of decrease of pressure of methane on the left side is given by a first-order process with respect to its pressure: $-\frac{dP_{CH_4}}{dt} = K_{CH_4} P_{CH_4}$. The problem states that the rate of diffusion is proportional to $\frac{P}{\sqrt{M}}$. Therefore, the rate constant $K$ is proportional to $\frac{1}{\sqrt{M}}$. So, $K = \frac{C}{\sqrt{M}}$ for some constant $C$. The formula given is $-\frac{dP}{dt} = k \frac{P}{\sqrt{M}}$. If we interpret this as $-\frac{dP}{dt} = \left(\frac{k}{\sqrt{M}}\right) P$, then the term in the parenthesis is the rate constant. Thus, the given value $2.5 \times 10^{-2}$ s$^{-1}$ is the rate constant for the diffusion of methane, i.e., $K_{CH_4} = 2.5 \times 10^{-2}$ s$^{-1}$.

The rate of decrease of the partial pressure of methane on the left side is given by: $-\frac{dP_{CH_4}}{dt} = K_{CH_4} P_{CH_4}$ where $P_{CH_4}$ is the partial pressure of methane on the left side at time $t$, and $K_{CH_4} = 2.5 \times 10^{-2}$ s$^{-1}$.

This is a first-order differential equation. Integrating this equation from time $t=0$ to $t$ and from initial pressure $P_{CH_4, 0}$ to $P_{CH_4}(t)$: $\int_{P_{CH_4, 0}}^{P_{CH_4}(t)} \frac{dP_{CH_4}}{P_{CH_4}} = \int_{0}^{t} -K_{CH_4} dt$ $[\ln P_{CH_4}]_{P_{CH_4, 0}}^{P_{CH_4}(t)} = -K_{CH_4} [t]_{0}^{t}$ $\ln P_{CH_4}(t) - \ln P_{CH_4, 0} = -K_{CH_4} t$ $\ln \left(\frac{P_{CH_4}(t)}{P_{CH_4, 0}}\right) = -K_{CH_4} t$

Initially, the pressure of methane on the left side is $P_{CH_4, 0} = 5.0$ atm. We want to find the time $t$ when the pressure of methane chamber drops to 4.0 atmospheres. Assuming this refers to the partial pressure of methane on the left side, $P_{CH_4}(t) = 4.0$ atm.

Substitute the values into the integrated equation: $\ln \left(\frac{4.0}{5.0}\right) = -(2.5 \times 10^{-2} \text{ s}^{-1}) t$ $\ln(0.8) = -(0.025 \text{ s}^{-1}) t$ $t = -\frac{\ln(0.8)}{0.025 \text{ s}^{-1}}$

Using the value $\ln(0.8) \approx -0.22314$: $t = -\frac{-0.22314}{0.025} \text{ s} = \frac{0.22314}{0.025} \text{ s} = 8.9256 \text{ s}$

We need to round the time to the nearest integer. $t \approx 9$ seconds.