Question

At what temperature is the root mean square speed of $N_2$ gas equal to that of propane gas at STP?

• A. ${173.7}^{\circ }\text{C}$
• B. $173.7\text{K}$
• C. $273\text{K}$
• D. $-{40}^{\circ }\text{C}$

Answer 1

The correct option is B $173.7\text{K}$

At STP, $T=273K,P=1bar$
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,
$v_{\text{rms}}\left(N_2\right)=v_{\text{rms}}\left(C_3{H}_8\right)$
$\sqrt{\frac{3RT}{28\times {10}^{-3}}}=\sqrt{\frac{3R\times 273}{44\times {10}^{-3}}}$
On solving, $T=173.7\text{K}$