Question

If the kinetic energy and RMS speed of a gas at a certain temperature are $4.0 \,kJ\, mol ^{-1}$ and $5.0 \times 10^{4}\, cm\, s ^{-1}$ respectively. The molecular weight of the gas is

• A. 16
• B. 32
• C. 64
• D. 44

Answer 1

Answer: (B)

First calculate temperature, using the average kinetic energy equation,

i.e. $KE =\frac{3}{2} R T$.

Then, calculate the molecular weight, using the formula for root mean square velocity,

i.e. $v_{ rms }=\sqrt{\frac{3 R T}{M}}$.

Given,

Kinetic energy $=4.0 \,kJ\, mol ^{-1} $

$ v_{ rms } =5.0 \times 10^{4} cm \,s ^{-1} $

or $ v_{ rms } =5.0 \times 10^{2} ms ^{-1} $

As we know that,

$KE =\frac{3}{2} R T $

$ 4 \times 1000 \,J / mol =\frac{3}{2} \times 8.314 \,J / K / mol \times T $

$ \therefore T=\frac{4 \times 1000 \times 2}{3 \times 8.314}=320 \,K $

Now,

$v_{ rms } =\sqrt{\frac{3 R T}{M}} $

$\therefore 5 \times 10^{2} m / s =\sqrt{\frac{3 \times 8.314\, J / K / mol \times 320\, K }{M}} $

$\left(\because 1 \,J =1 \,kg \,m ^{2} s ^{-2}\right) $

$\therefore M= \frac{3 \times 8.314 \times 320}{25 \times 10^{4}} kg / mol $

$= 0.03192 \,kg / mol =31.92 \approx 32\, g / mol$

Hence, the molecular weight of the gas in $32\, g / mol$.