Question

If the kinetic energy and RMS speed of a gas at a certain temperature are $4.0kJmo{l}^{-1}$ 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. $KE=\frac{3}{2}RT$.

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

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

Given,

Kinetic energy $=4.0kJmo{l}^{-1}$

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

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

As we know that,

$KE=\frac{3}{2}RT$

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

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

Now,

$v_{rms}=\sqrt{\frac{3RT}{M}}$

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

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

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

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

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