What is the speed of sound in a monoatomic ideal gas contained within a box whose volume is 2.5 cube meter?

Question

(y=1.67)The pressure of the gas is 3.5 x10fifth Pa. The total mass of gas is 2.3kg.

Answer 1

In an ideal monoatomic gas, the speed of sound is:

v = sqrt(gamma*k*T/m)

where gamma = 5/3 for a monoatomic gas.

Since
PV = NkT

where:
- the number of atoms = N
- k is Boltzmann's constant
- T is temperature in Kelvin

3.5e5 * 2.5 = 8.75e5 = NkT

2.3 = total mass = N*m

Therefore,
8.75e5/2.3 = (NkT)/(Nm) = kT/m
so
kT/m = 8.75e5/2.3 = 3.8e5

gamma*(kT/m) = (5/3)*(3.8e5)
= 6.34e5

So
v = sqrt(gamma*(kT/m)) = sqrt(6.34e5) = 796.3 m/s

Answer 2

The speed of sound in a gas (which is also the mean molecular velocity) is found using this simple relation:

a = sqrt(gamma*R*T / M)

where gamma = (f + 2) / f = 5/3 for helium, f is the number of degrees of freedom, R is the gas constant, T is the temperature of the gas, and M is the gas's molar mass.