Question

A 0.75kg sample of gaseous monatomic Helium (He) has an rms velocity of 2500m/s in an insulated container of 0.8m3.

a) What is the pressure of the gas? (3.5pts)

b) What is the temperature of the gas? (3pts)

Answer 1

Part A.

We know that rms velocity of gaseous molecule is given by:

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

Given that V_rms = 2500 m/s, Mw = molecular weight of He = 4.00 gm/mol = 4.00*10^-3 kg/mol

So,

T = Mw*V_rms^2/(3*R)

R = gas constant = 8.314, So

T = 4.00*10^-3*2500^2/(3*8.314)

T = 1002.32 K = temperature of the gas

Now using ideal gas law:

PV = nRT

P = n*R*T/V

V = Volume of gas = 0.8 m^3

n = number of moles = m/Mw = 0.75 kg/(4.00*10^-3 kg/mol) = 187.5 moles

So,

P = 187.5*8.314*1002.32/0.8

P = 1.953*10^6 N/m^2 = Pressure of the gas

Part B.

T = 1002.32 K = temperature of the gas (1002.32 K = 729.17 C)

Let me know if you've any query.