Question

3. A 50.0 L sample of gas collected in the upper atmosphere at a pressure of 18.3 mmHg is compressed into a 150.0 mL container at the same temperature.



a. What is the new pressure, in atm?



b. To what volume would the original sample have had to be compressed to exert a pressure of 10.0 atm?

Answer 1

Answer:

a. $P_2=8.03atm$

b. $V_2=0.024L=24mL$

Explanation:

Hello,

a. In this case, we use the Boyle's law as en inversely proportional relationship between pressure and volume:

$P_1V_1=P_2V_2$

Thus, for the given conditions, one computes the new pressure as shown below:

$P_2=\frac{P_1V_1}{V_2} =\frac{18.3mmHg*50.0L}{150.0mL*\frac{1L}{1000mL} }*\frac{1atm}{760mmHg} \\ \\P_2=8.03atm$

b. Now, we should find the final volume for a new pressure of 10 atm:

$V_2=\frac{P_1V_1}{P_2} =\frac{50.0L*18.3mmHg*\frac{1atm}{760mmHg} }{50atm}\\ \\V_2=0.024L=24mL$

Best regards.