Question 811: Thermodynamics & KTG

Question

At S.T.P., the mean free path of gas molecule is 1500 d , where ' $d$ ' is diameter of molecule. What will be the mean free path at 373 K at constant volume?

Accepted Answer

The mean free path of gas molecules at standard temperature and pressure (S.T.P.) is given by:
$$ \lambda_0 = 1500d $$
where $ \lambda_0 $ is the mean free path at S.T.P., and $ d $ is the diameter of a gas molecule.
At constant volume, the mean free path $\lambda$ is affected by temperature according to the relationship:
$$ \lambda \propto \frac{T}{P} $$
Since the pressure remains constant (as volume is constant), the relationship simplifies to:
$$ \lambda \propto T $$
Thus, to find the mean free path at a new temperature $ T_1 = 373 \, 	ext{K} $, we use the ratio of the temperatures:
$$ \lambda_1 = \left(\frac{T_1}{T_0}ight) \lambda_0 $$
Substituting the known temperatures:
$ T_0 = 273 \, 	ext{K} $ (S.T.P.)
$ T_1 = 373 \, 	ext{K} $
The mean free path at 373 K becomes:
$$ \lambda_1 = \left(\frac{373}{273}ight) 	imes 1500d $$
Thus, the correct answer is Option B:
$ \frac{373}{273} 	imes 1500 \mathrm{~d} $

Answer

1500 d

Answer

$\frac{373}{273} 	imes 1500 \mathrm{~d}$

Answer

$\frac{273}{373} 	imes 1500 \mathrm{~d}$

Answer

$\sqrt{\frac{373}{273}} 	imes 1500 \mathrm{~d}$

Thermodynamics & KTG

At S.T.P., the mean free path of gas molecule is 1500 d , where ' d ' is diameter of molecule. What will be the mean free path at 373 K at constant volume?

Choose the correct answer: