Question

A 8.7 L (8.7x10-3 m3) vessel contains 4.3 moles of ideal gas at a pressure of 2.1X106 Pa.

Find

(a) the temperature of the gas. (R = 8.314 J/mol·K)

(b) the average kinetic energy of a gas molecule in the vessel

(c) the rms speed if the gas is O₂ (M = 32.0x10^-3 kg/mol)

Answer 1

Part A.

Using ideal gas law:

PV = nRT

T = P*V/(n*R)

P = Pressure = 2.1*10^6 Pa

V = 8.7 L = 8.7*10^-3 m^3

n = number of moles = 4.3 moles

So,

T = 2.1*10^6*8.7*10^-3/(4.3*8.314)

T = 511.0 K = temperature of gas

2.

Average kinetic energy of molecules is given by:

KE_avg = (3/2)*k*T

k = Boltzmann's constant = 1.38*10^-23

T = temperature of gas = 511.0 K

So,

KE_avg = 3*1.38*10^-23*511.0/2

KE_avg = 1.06*10^-20 J

3.

rms speed of gas molecules is given by:

V_rms = sqrt (3*R*T/Mw)

Mw = molecular weight of oxygen = 32.0*10^-3 kg/mol

So,

V_rms = sqrt (3*8.314*511.0/(32.0*10^-3))

V_rms = 631.1 m/s = rms speed

Let me know if you've any query.