First slide

Maxwell speed distribution law

Question

Two vessels A and B contain ideal gases with the temperature of B double that of A. Both gases are heated, so that they attain the same temperature. It is found that the fractional increase in the most probable speed of gas in vessel ,A is double that of the mean speed of gas in B. The ratio of the final to the initial temperature of gas in vessell is

Moderate

A

$3 - 2 \sqrt{2}$
B

$2 - 3 \sqrt{2}$
C

3+ $2 \sqrt{2}$
D

$2 + 3 \sqrt{2}$

Solution

$E_{A} = \frac{3}{2} {kT}_{A}; E_{B} = \frac{3}{2} k (2 T_{A})$
and ${E_{A}}^{'} = \frac{3}{2} kT; {E_{B}}^{'} = \frac{3}{2} kT$
$\frac{∆ v_{p}}{v_{p}} = (\sqrt{\frac{T}{T_{A}}} - 1) = 2 \frac{∆ v_{avg}}{v_{avg}} = 2 (\sqrt{\frac{T}{2 T_{A}}} - 1)$
Let $\frac{T}{T_{A}} = x, the above equation reduces to$
$(\sqrt{x} - 1) = 2 (\sqrt{\frac{x}{2}} - 1) or \sqrt{x} (\sqrt{2} - 1) = 1$
or $x = [\frac{1}{(3 - 2 \sqrt{2})}]$
or x = 3 $+ 2 \sqrt{2}$

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