Question

Gas Laws

Units in Gas Law Problems

One problem associated with gas law problems is unit agreement. It is important that pressure, volume and temperature units match or agree within a problem. Solve 1-4 using the factor-label method. Problems 5 and 6 are done without factor-label.

1. Convert 1.52 atm to kPa 2. Convert 85 kPa to mmHg

3. Convert 156.3 mmHg to atm 4. Convert 950 torr to kPa

5. Convert -250.0 °C to K 6. Convert 253 K to °C

Boyle’s Law

7. Equation for Boyle’s Law:

8. A gas occupies 12.3 L at 825.7 mmHg. What will the pressure be when the volume is 75 L?

9. A gas occupies 25 L at 2.5 atm. What is the volume if the pressure changes to 1.5 atm?

10. You have 50 L of CO₂ gas at standard temperature and pressure (STP). What would need to be done to the pressure to cut the volume of gas in half?

Charles’ Law

11. Equation for Charles’ Law:

12. How do these two factors relate to each other?

13. What does temperature actually measure?

14. At 27.8 °C, a gas occupies 1500 m. What volume will it have at 100.0 °C?

15. What temperature (in K) must a gas be if it occupied 1.396 L at 72.3 °C and now occupies 1.044 L?

Gay-Lussac’s Law

16. Equation for Charles’ Law:

17. A gas cylinder contains nitrogen gas at 10-atm pressure and a temperature of 20˚C. The cylinder is left in the sun, and the temperature of the gas increases to 50˚C. What is the pressure in the cylinder?

18. A bike tire has a volume of 0.850L at a pressure of 40 psi and 0˚C. What will be the pressure of the tire at 35˚C?

19. An aerosol can has a fixed volume of gas at 4.0-atm of pressure and room temperature (25˚C). If the pressure inside the can reaches 5.9-atm the can will explode. The can is thrown into a fire that is 400˚C. Will the can explode? Show all calculations to support your answer.

Answer 1

1. 1.52 atm = [1.52 atm / 1 ] x [101.325 kpa / 1 atm] =154.014 kpa

2. 85 kpa =[ 85 kpa / 1] x [ 7.5 mm Hg / 1 kpa] = 637.5 mm Hg

3. 156.3 mm Hg = [156.3 mm Hg / 1] x [ 1 atm / 760 mm Hg] = 0.2057 atm

4. 950 Torr = [950 Torr / 1] x [ 1 kpa / 7.5 Torr ] = 126.7 kpa

5. 250 ⁰C = [250 + 273 ] K = 523 K

6. 253 K = [253 - 273 ] ⁰C = -20 ⁰C

7. Boyle's law equation, PV = constant or p₁V₁ = P₂V₂ at constant temperature.

8. Let P₂ be the pressure at volume V₂ = 75 L

therefore

9. Let V₂ be the volume at pressure P₂ = 1.5 atm

therefore V₂ = P₁V₁ / P₂

                    = [ 2.5 atm x 25 L]/ 1.5 atm

                  = 41.7 L

10. Pressure has to be increased to reduce the volume.

Given P₁ = 1 atm [ STP condition is 1 atm pressure and 273 K] and V₁ = 50 L

        P₂ = ? and V₂ = 1/2 V₁ = 50 /2 = 25 L

        

P₂ = P₁V₁/ V₂

                  = [1 atm x 50 L ] / 25 L

                  = 2 atm.

Thus the pressure has to be doubled.

11. V/T is a constant or V₁/T₁ = V₂/T₂ at constant pressure

12. Volume is directly proportional to Temperature. As the volume increases temperature also increases.

13. Temperature actually measures the average kinetic energy of partlcles of the substance.

14. Let V₂ be the volume at temperature T₂ = 100 ⁰C = 100 + 273 = 373 K

     T₁ = 27.8 ⁰C = 273 + 27.8 = 300.8 K , V₁ = 1500 m³

therefore V₂ = T₂ V₁ / T₁ = 373 x 1500 / 300.8 = 1860 m³

15. Let T₂ be the temperature of the gas occupying a volume of 1.044 L = V₂

Given the volume is 1.396 L at 72.3 ⁰ C = 273 + 72.3 = 345.3 K

T₂ = V₂ x T₁ / V₁ = 1.044 x 345.3/ 1.396 = 258.2 K

16. P / T is a constant or P₁/T₁ = P₂/T₂ is the equation for Lussac's law. at constant volume

17. Let P₂ be the pressure at temperature T₂ = 50 ⁰C = 273 + 50 = 323 K

     P₁ = 10 atm   T₁ = 20 ⁰C = 273 + 20 = 293 K

therefore P₂ = T₂ x P₁/T₁

                  = 323 x 10 / 293 = 11 atm

18.

Let P₂ be the pressure at temperature T₂ = 35 ⁰C = 273 + 35 = 308 K

     P₁ = 40 psi   T₁ =0 ⁰C = 273 K

therefore P₂ = T₂ x P₁/T₁

                  = 308 x 40 / 273 = 45 psi

19. .

Let P₂ be the pressure at temperature T₂ = 400 ⁰C = 273 + 400= 673 K

     P₁ = 4 atm   T₁ = 25 ⁰C = 298 K

therefore P₂ = T₂ x P₁/T₁

                  = 673 x 4 / 298 = 9 atm

Since the final pressure is 9 atm which is higher than 5.9. Hence the can will explode.