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Question

Chemistry Question on intermolecular forces

The average molecular speed is greatest in which of the following gas samples ?

A

1.0 mole N2N_2 at 560 K

B

0.50 mole of Neat 500 K

C

0.20 mole CO2CO_2 at 440 K

D

2.0 mole of He at 140 K

Answer

2.0 mole of He at 140 K

Explanation

Solution

υˉ(Ne)=(8R×560π×28)12\bar{\upsilon} (Ne) = \left( \frac{8R \times 560}{\pi \times 28} \right) ^{\frac{1}{2}} (A) υˉ(N2)=(8R×560π×28)12=(160Rπ)12\bar{\upsilon} (N_2) = \left( \frac{8R \times 560}{\pi \times 28} \right) ^{\frac{1}{2}} = \left( \frac{160 \, R}{\pi}\right)^{\frac{1}{2}} (B) υˉ(Ne)=(8R×500π×20)12=(200Rπ)12\bar{\upsilon} (Ne) = \left( \frac{8R \times 500}{\pi \times 20} \right) ^{\frac{1}{2}} = \left( \frac{200 \, R}{\pi}\right)^{\frac{1}{2}} (C) υˉ(CO2)=(8R×440π×44)12=(80Rπ)12\bar{\upsilon} (CO_2) = \left( \frac{8R \times 440}{\pi \times 44} \right) ^{\frac{1}{2}} = \left( \frac{80 \, R}{\pi}\right)^{\frac{1}{2}} (D) υˉ(He)=(8R×140π×4)12=(280Rπ)12\bar{\upsilon} (He) = \left( \frac{8R \times 140}{\pi \times 4} \right) ^{\frac{1}{2}} = \left( \frac{280 \, R}{\pi}\right)^{\frac{1}{2}} Thus, υ(He)\upsilon(He) is maximum .