Question

The pressure in a bulb dropped from 2000 to 1500 mm Hg in 47 min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1:1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 min.

• A. $\frac{M_{gas}}{M_{O_2}}=0.666$
• B. $\frac{M_{gas}}{M_{O_2}}=1.236$
• C. $\frac{M_{gas}}{M_{O_2}}=2.472$
• D. $\frac{M_{gas}}{M_{O_2}}=5.236$

Answer 1

Answer: (B)

$\frac{M_{gas}}{M_{O_2}}=1.236$

The change of pressure of oxygen in 47 min is 500 mm Hg. The change of pressure of oxygen after 74 min is

$\frac{500}{47}\times 74=787.2mm$

In the 1:1 molar ratio mixture of oxygen and another gas, each of them has an equal pressure of 2000 mm because the total pressure is given to be 4000 mm Hg.

The pressure of oxygen left after 74 min is

$2000-787.2=1212.8mmHg$

$\frac{r_{gas}}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$(Graham's law of diffusion)

$\frac{V_{gas}\left(\text{diffused}\right)}{V_{O_2}\left(\text{diffused}\right)}\times \frac{t_{O_2}}{t_{gas}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

$\frac{P_{gas}\left(\text{diffused}\right)}{P_{O_2}\left(\text{diffused}\right)}\times \frac{t_{O_2}}{t_{gas}}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

Both diffuse for the same time, so

$\frac{P_{gas}\left(\text{diffused}\right)}{P_{O_2}\left(\text{diffused}\right)}=\sqrt{\frac{M_{O_2}}{M_{gas}}}$

or $\frac{P_{gas}\left(\text{diffused}\right)}{787.2}=\sqrt{\frac{32}{79}}$

$P_{gas}\left(\text{diffused}\right)=500.8mm$

The pressure of gas left after 74 min is

$2000-500.8=1499.2mmHg$

$\text{Molar ratio of the gas and oxygen left}=\frac{P_{(\text{gas)}}\text{left}}{P_{\left(O_2\right)}\text{left}}=\frac{M_{gas}}{M_{O_2}}=\frac{1499.2}{1212.8}=1.236$

Hence, the correct option is $\text{B}$

$\frac{M_{gas}}{M_{O_2}}=1.236$