Question

Consider two 1-L samples of gas, one H2 and one O2, both at 1 atm and 25°C. Which statement most accuratey compares the gases? A. The rate of effusion of the O2 is greater than that of the H2. B. The rate of effusion of the O2 is equal to that of the H2. C. The average molecular speed of the O2 is greater than that of the H2. D. The average molecular speed of the O2 is equal to that of the H2.

Answer 1

Answer: none of these

Explanation:-

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = 1 L

T= Temperature of the gas = 25°C = 298 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

$n=\frac{PV}{RT}=\frac{1\times 1}0.0821\times 298}=0.04moles$

1 mole of oxygen weigh = 32 g

0.04 moles of oxygen weigh =$\frac{32}{1}\times 0.04=1.28g$

1 mole of hydrogen weigh = 2 g

0.04 moles of oxygen weigh =$\frac{2}{1}\times 0.04=0.08g$

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{H_2}}{Rate_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$

$\frac{Rate_{H_2}}{Rate_{O_2}}=\sqrt{\frac{1.28}{0.08}$

$\frac{Rate_{H_2}}=4\times {Rate_{O_2}}$

Thus rate of effusion of hydrogen is 4 times faster than rate of effusion of oxygen.

The formula for average speed is  

$\nu_{av}=\sqrt{\frac{8RT}{\pi M}}$

$\text{Average speed}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$

$\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{1.28}{0.08}$

$\nu_{H_2}=4\times {\nu_{O_2}}$

Thus average speed of hydrogen is 4 times faster than average speed of oxygen.

Thus none of the given options is correct.