Question

At what temperature is the root mean square speed of $N_2$ gas equal to that of propane gas at $440\text{K}$?

• A. $280\text{K}$
• B. $300\text{K}$
• C. $320\text{K}$
• D. $260\text{K}$

Answer 1

The correct option is A $280\text{K}$

For propane, $T=440K,P=1bar$
molar mass of $N_2$ = 28 g = $28\times {10}^{-3}$ kg/mol
molar mass of $C_3{H}_8$ = 44 g = $44\times {10}^{-3}$ kg/mol
Hence, putting values,
$v_{\text{rms}}\left(N_2\right)=v_{\text{rms}}\left(C_3{H}_8\right)$
$\sqrt{\frac{3RT}{28\times {10}^{-3}}}=\sqrt{\frac{3R\times 440}{44\times {10}^{-3}}}$
On solving, $T=280\text{K}$