A $0.50 \mathrm {~m} ^ { 3 }$ gas tank holds 3.0 moles of ideal diatomic nitrogen gas at a temperature of 350 K. The atomic mass of nitrogen is 14.0 g/mol. What is the rms speed of the molecules? (The Boltzmann constant is $\left. 1.38 \times 10 ^ { - 23 } \mathrm {~J} / \mathrm { K } , N _ { \mathrm { A } } = 6.022 \times 10 ^ { 23 } \mathrm { molecules } / \mathrm { mol } . \right)$
A) 97
B) 21
C) 790
D) 390
E) 560

Question

A $0.50 \mathrm {~m} ^ { 3 }$ gas tank holds  3.0  moles of ideal diatomic nitrogen gas at a temperature of  350 K. The atomic mass of nitrogen is  14.0 g/mol. What is the rms speed of the molecules? (The Boltzmann constant is $\left. 1.38 \times 10 ^ { - 23 } \mathrm {~J} / \mathrm { K } , N _ { \mathrm { A } } = 6.022 \times 10 ^ { 23 } \mathrm { molecules } / \mathrm { mol } . \right)$
A) 97
B) 21
C) 790
D) 390
E) 560

Quizplus · Accepted Answer

The rms speed $ v_{rms} $ is given by the formula $ v_{rms} = \sqrt{\frac{3kT}{m}} $, where $ k $ is the Boltzmann constant, $ T $ is the temperature, and $ m $ is the mass of one molecule. For diatomic nitrogen, the molar mass is $ 28 \, \text{g/mol} $ or $ 0.028 \, \text{kg/mol} $. The mass of one molecule is $ \frac{0.028}{6.022 \times 10^{23}} \, \text{kg} $. Substituting these values into the formula gives the rms speed as approximately 560 m/s.

A 0.50 m30.50 \mathrm {~m} ^ { 3 }0.50 m3 Gas Tank Holds 3

A $0.50 \mathrm {~m} ^ { 3 }$ Gas Tank Holds 3