Respuesta :
Explanation:
Let the temperature at which gas is moving be T
Pressure exerted by the gas , P = 4.35 atm
Volume occupied by the gas ,V = 3.5 l
Moles of bromine gas,n = 1.4 moles
Consider the bromine gas is behaving ideal .
[tex]PV=nRT[/tex]
[tex]4.35 atm\times 3.5 L=1.4mol\times 0.0820 L atm/K mol\times T[/tex]
[tex]T=\frac{4.35 atm\times 3.5 L}{1.4mol\times 0.0820 L atm/K mol}=132.62 K[/tex]
Mass of bromine gas,M = Moles × Molar mass = 1.4 mole × 160 g/mol = 224 g = 0.224 kg
Value of R = 8.314 J/mol K
Root mean square velocity of the bromine gas:
[tex]\mu_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314 J/mol K\times 132.62 K}{0.224 kg}}=121.51 m/s[/tex]
Average velocity of the bromine gas:
[tex]v_{avg}=\sqrt{\frac{8RT}{\piM}}=\sqrt{\frac{8\times 8.314 J/mol K\times 132.62 K}{3.14\times 0.224 kg}}=111.98 m/s[/tex]
Most probable velocity:
[tex]v_{m}=\sqrt{\frac{2RT}{M}}=\sqrt{\frac{2\times 8.314 J/mol K\times 132.62 K}{0.224 kg}}=99.22 m/s[/tex]
