Question

A volume of nitrogen gas has a density of 1.301.30 kg/m3 and a pressure of 1.151.15 atm. Find the temperature of the gas and the rms speed of its molecules.

Answer 1

Answer :

The temperature of the gas is, 301.7 K

The root mean square speed is, $5.18\times 10^{2}m/s$

Explanation :

To calculate the volume of argon gas we are using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\frac{\rho RT}{M}$

where,

P = pressure of nitrogen gas = 1.15 atm

V = volume of nitrogen gas

T = temperature of nitrogen gas

R = gas constant = 0.0821 L.atm/mole.K

w = mass of nitrogen gas

M = molar mass of nitrogen gas = 28 g/mole

$\rho$ = density of nitrogen gas = $1.30kg/m^3=1.30g/L$

Now put all the given values in the ideal gas equation, we get:

$1.15atm=\frac{(1.30g/L)\times (0.0821L.atm/mole.K)\times T}{28g/mol}$

$T=301.7K$

Therefore, the temperature of the gas is, 301.7 K

Now we have to determine the root mean square speed of the molecule.

The formula used for root mean square speed is:

$\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}$

where,

$\nu_{rms}$ = root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature = 301.7 K

M = atomic mass of nitrogen gas = 0.028 kg/mole

$N_A$ = Avogadro’s number = $6.02\times 10^{23}mol^{-1}$

Now put all the given values in the above root mean square speed formula, we get:

$\nu_{rms}=\sqrt{\frac{3\times (1.38\times 10^{-23}J/K)\times (6.02\times 10^{23}mol^{-1})\times (301.7K)}{0.028kg/mol}}$

$\nu_{rms}=5.18\times 10^{2}m/s$

The root mean square speed is, $5.18\times 10^{2}m/s$