A tank holds 184 moles of argon gas (atomic mass = 6.633521 × 10 -26 kg) at a pr

Question

A tank holds 184 moles of argon gas (atomic mass = 6.633521 × 10-26 kg) at a pressure of 101300 Pa and a temperature of 293 K. Recall that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas

1. What is the volume of the tank?

2. How much kinetic energy is contained in this gas?

3. What is the average speed of one of the argon atoms?

4. The escape speed of Earth is 11200 m/s. What temperature would the argon gas need to have so that the average speed of the gas atoms was equal to this escape speed?

Explanation / Answer

1) pV = nRT
101300Pa * V = 184mol * 8.314J/K·mol * 293K
V = 4.424 m³ = 4424 L

2) KE = (3/2)nRT = (3/2) * 184mol * 8.314J/K·mol * 293K = 672 kJ

3) Vrms = (3RT / M)
Molar mass M = 6.633521e-26kg/atom * 6.022e23atom/mol = 0.0399 kg/mol
so
Vrms = (3 * 8.314J/K·mol * 293K / 0.0399kg/mol) = 427 m/s

4) 11200 m/s = (3 * 8.314J/K·mol * T / 0.0399kg/mol)
solves to
T = 2.01*10^5 K

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