Question:
At \(300\,\mathrm{K}\), the diffusion rate of one mole of an ideal gas is \(0.082\,\mathrm{L\,s^{-1}}\). What is the pressure (in atm) of this gas which can diffuse in \(100\,\mathrm{s}\)?
\[
\left(R=0.082\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}\right)
\]
At \(300\,\mathrm{K}\), the diffusion rate of one mole of an ideal gas is \(0.082\,\mathrm{L\,s^{-1}}\). What is the pressure (in atm) of this gas which can diffuse in \(100\,\mathrm{s}\)?
\[
\left(R=0.082\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}\right)
\]
Show Hint
For one mole of an ideal gas,
\[
\boxed{
PV=RT.
}
\]
Always calculate the volume first, then substitute into the ideal gas equation.
Updated On: Jul 15, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Calculate the volume occupied by one mole of gas.
Given diffusion rate,
\[
0.082\,\mathrm{L\,s^{-1}}.
\]
In \(100\,\mathrm{s}\),
\[
V
=
0.082\times100
=
8.2\,\mathrm{L}.
\]
Step 2: Apply the ideal gas equation. Using \[ PV=nRT, \] where \[ n=1,\qquad T=300\,\mathrm{K}, \] \[ P = \frac{nRT}{V} = \frac{1\times0.082\times300}{8.2} = 3\,\mathrm{atm}. \] Hence, \[ \boxed{3\,\mathrm{atm}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
Step 2: Apply the ideal gas equation. Using \[ PV=nRT, \] where \[ n=1,\qquad T=300\,\mathrm{K}, \] \[ P = \frac{nRT}{V} = \frac{1\times0.082\times300}{8.2} = 3\,\mathrm{atm}. \] Hence, \[ \boxed{3\,\mathrm{atm}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
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