Respuesta :
Answer :
The temperature of the gas is, 301.7 K
The root mean square speed is, [tex]5.18\times 10^{2}m/s[/tex]
Explanation :
To calculate the volume of argon gas we are using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\frac{\rho RT}{M}[/tex]
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
[tex]\rho[/tex] = density of nitrogen gas = [tex]1.30kg/m^3=1.30g/L[/tex]
Now put all the given values in the ideal gas equation, we get:
[tex]1.15atm=\frac{(1.30g/L)\times (0.0821L.atm/mole.K)\times T}{28g/mol}[/tex]
[tex]T=301.7K[/tex]
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:
[tex]\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}[/tex]
where,
[tex]\nu_{rms}[/tex] = root mean square speed
k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]
T = temperature = 301.7 K
M = atomic mass of nitrogen gas = 0.028 kg/mole
[tex]N_A[/tex] = Avogadro’s number = [tex]6.02\times 10^{23}mol^{-1}[/tex]
Now put all the given values in the above root mean square speed formula, we get:
[tex]\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}}[/tex]
[tex]\nu_{rms}=5.18\times 10^{2}m/s[/tex]
The root mean square speed is, [tex]5.18\times 10^{2}m/s[/tex]
