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devlian [24]
3 years ago
9

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 a

t 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?
Chemistry
1 answer:
pochemuha3 years ago
4 0

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.

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