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

a.
`P_2=8.03atm`

b.
`V_2=0.024L=24mL`

Step-by-step 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=(P_1V_1)/(V_2) =(18.3mmHg*50.0L)/(150.0mL*(1L)/(1000mL) )*(1atm)/(760mmHg) \\ \\P_2=8.03atm`

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

`V_2=(P_1V_1)/(P_2) =(50.0L*18.3mmHg*(1atm)/(760mmHg) )/(50atm)\\ \\V_2=0.024L=24mL`

Best regards.