Gas Law Problems (1 number still blank; the bold one is the answer)

Kinetic Theory and Pressure conversions:

For questions 1-5, please answer in full sentences.

1. If you inflate a raft, what will happen to the pressure of the raft as you add more gas? the pressure inside the raft will increase
2.  
1. Why does this occur?

By adding gas, the numbers of particles increase. There are collisions of billions of rapidly moving particles in a gas inside walls of the raft, as the result of pressure that is exerted by the gas. With increasing the number of particles, the number of collisions is increasing as well. Therefore, the gas pressure increase

3. If you heat up an aerosol can, what will happen to the can? Aerosol can may explode when heated
1. Why does this occur? As gas inside the can is heated, the temperature increases that allow the kinetic energy of the particles also increasing.  So it is causing more collisions, thus more pressure. And also gas in sealed container may generate massive pressure when heated. Therefore, Aerosol can may explode when heated
4. When you sit on an exercise ball, what will happen to the pressure inside the ball? The pressure inside the exercise ball will increase

a.      Why does this occur? When a person sit on a ball, the particles in gas are forced closer together or compressed. When the gas is compressed, the kinetic energy of molecules will increase as they are forced closer together. Due to the speed of a gas molecules increase, the gas molecules hit the container more often. Therefore, the pressure increases.

5. Use kinetic molecular theory to explain
1. Why evaporation causes a decrease in a liquid’s temperature

Because evaporation is a process that allows liquid to change state to a gas and evaporation requires energy to occur. So the molecules that evaporate are those with the kinetic energy. Then after the evaporation, the energy of the remaining liquid molecules is lower, that makes the temperature of the liquid decreasing.

2. Why gases are more easily compressed than solids and liquids

Because the particle of gas are too far apart to be attracted to each other and no strong forces of attraction that hold them together unlike liquid where the molecules are closely packed and they cannot move freely, however can still move around because the forces of attraction between them is not as weak as gas but not as strong as solid, and solids, where the molecules are closely packed and arrange in order that makes the molecules in their fixed positions. Therefore, gas molecules can move freely and compressed into the free space between them

6. A small sample of gas is released in a corner of the room and starts to diffuse to the other side.  If the room pressure is increased, will the gas diffuse faster, slower, or at the same speed?  Explain.
7. For each of the following, convert from the units given to the desired units using dimensional analysis.  SHOW YOUR WORK if you wish to receive credit.  Include all units and box or highlight your answers.
1. 202.6 kPa = ? atm

Given: pressure= 202.6 kPa; 1 atm = 101.3 kPa

Required: pressure in atm

Solve: 202.6 kPa x 1 atm/101.3 kPa = 2 atm

Paraphrase: The pressure is 2 atm

Evaluate:  The result makes sense because the conversion factor is much less than 1, so the values expressed in atm is smaller than the value expressed in kPa

2. 560 kPa = ? mmHg

Given: pressure = 560 kPa; 1 atm = 760 mmHg; 1 atm = 101.3 kPa

Required: pressure in mmHg

Solve: 560 kPa x 760 mmHg/101.3 kPa = 4201.38 mmHg

Paraphrase: The pressure is 4201.38 mmHg

Evaluate: The result makes sense because the conversion factor is greater than 1, so the value expressed in mmHg is bigger than the value expressed in kPa

3. 5 atm = ? kPa

Given: pressure= 5 atm; 1 atm = 101.3 kPa

Required: pressure in kPa

Solve: 5 atm x 101.3 kPa/ 1 atm = 506.5 kPa

Paraphrase: The pressure is 506.5 kPa

Evaluate: The result makes sense because the conversion factor is much greater than 1, so the value expressed in kPa is bigger than the value expressed in atm

4. 3 atm = ? mmHg

Given: pressure= 3 atm; 1 atm= 760 mmHg

Required: Pressure in mmHg

Solve: 3 atm x 760 mmHg /1 atm = 2280 mmHg

Pressure: The pressure is 2280 mmHg

Evaluate: The result makes sense because the conversion factor is much greater than 1, so the value expressed in mmHg is bigger than the value expressed in atm

5. 830 mmHg = ? kPa

Given: pressure= 830 mmHg; 1 atm= 101.3 kPa; 1 atm= 760 mmHg

Required: pressure in kPa

Solve: 830 mmHg x 101.3 kPa /760 mmHg = 110.63 mmHg

Paraphrase: The pressure is 110.63 kPa

Evaluate: The result makes sense because the conversion factor is much less than 1, so the values expressed in kPa is smaller than the value expressed in mmHg

6. 43 mmHg = ? atm

Given: pressure= 43 mmHg; 1 atm = 760 mmHg

Required: pressure in atm

Solve: 43 mmHg x 1 atm / 760 mmHg = 0.057 atm

Paraphrase: The pressure is 0.057 atm

Evaluate: The result makes sense because the conversion factor is much less than 1, so the values expressed in atm is smaller than the value expressed in mmHg

Boyle’s Law
1. What is the new volume when a  125 mL container at 120.0 kPa is expanded until the pressure is 60.0 kPa? (assume the temperature is constant)

Given: V₁= 125 mL; P₁= 120.0 kPa; P₂= 60 kPa                                                                                                                                                                        Required: V₂                                                                                                                                                                                                                              Solve: V₂= P₁ x V₁ / P₂                                                                                                                                                                                                                   V₂  = 125 mL x 120 kPa/ 60 kPa = 250 mL

Paraphrase: The new volume is 250 mL

Evaluate: A decrease in pressure at constant temperature must correspond to a proportional decrease in volume. The calculated result agrees with both kinetic theory is expressed to the proper number of significant figures

2. What is the new pressure if a 100.0 L  container at standard pressure (101.3 kPa) is compressed until the volume is 50.0 L? (Assume the temperature is constant)

Given: V₁= 100L; P₁= 101.3 kPa; V₂= 50 L

Required: P₂

Solve: P₂= V₁ x P₁ / V₂

P₂ = 100 L x 101.3 kPa / 50 L = 202.6 kPa

Paraphrase: The new pressure is 202.6 kPa

Evaluated: A decrease in volume at constant temperature must correspond to a proportional decrease in pressure. The calculated result agrees with both kintic theory is expressed to the proper number of significant figures

Charles’s Law
3. What is the new volume of a 10.0 mL container at 0.00 K when the temperature is adjusted to 283 K? (Assume pressure is constant)

Given: V₁= 10 mL; T₁= 0 K; T₂= 283K

Required: V₂

Solve: V₂= V₁ x T₂ / T₁

V₂ = 10 mL x 283 K /  0 K = undefined

Paraphrase: The new volume is undefined

Evaluate: The temperature increase as the pressure decrease. This result agrees with both kinetic theory and Charles’s law

4. A 50.0 mL container is at 273 K. After the volume is adjusted to 100.0 mL, what is the new temperature? (Assume pressure is constant)

Given: V₁=50 mL; T₁= 273K; V₂= 100mL

Required: T₂

Solve: T₂= V₂ x T₁ / V₁

T₂= 100 mL x 273 K / 50 mL = 546 K

Paraphrase: The new temperature is 546 K
Evaluate: The temperature increase as the volume increase. This result agrees with both kinetic theory and Charles’s law

Gay-Lussac’s Law
5. What is the new pressure of a set volume of gas at 101 kPa when it is heated from 295 K to 400K? (assume that volume is constant)

Given: P1= 101 kPa; T1= 295K; T2= 400K

Required: P2

Solve: P2= P1 x T2 / T1

P₂ = 101 kPa x 400 K / 295 K = 136.95 kPa

Paraphrase: The new pressure is 136.95 kPa

Evaluate: From the kinetic theory, the expected outcome if the temperature increase, the pressure is also increasing if the volume remains constant. The calculated value does show such an increase.
6.  A container of gas at 325K and 500 kPa decreases in pressure to 50 kPa.  What is the new temperature of the gas? (assume that volume is constant)

Given: T1: 325K; P1= 500kPa; P2= 50 kPa

Required: T2

Solve: T2: P2x T1/ P1

T₂ = 50 kPa x 325 K / 500 kPa = 32.5 K

Paraphrase: The new temperature is 32.5 K

Evaluate: From the kinetic theory, the expected outcome if the pressure decrease, the temperature is also decreasing if the volume remains constant. The calculated value does show such an decrease
Combined Gas Law
7. What is the new pressure of a 10.5 L sample of gas at 350 K and 101 kPa when it is heated to 375 K and the volume increases to 11.5L?  

Given: V₁= 10.5 L; T₁= 350K; P₁= 101 kPa; T₂= 375K; V₂=11.5 L

Required: P₂

Solve: P₂= P₁ x T₂ x V₁ / T₁ x V₂

P₂ = 101 kPa x 375 K x 10.5 L / 350 K x 11.5 L = 98.80 kPa

Paraphrase: The new pressure is  98.80 kPa

Evaluate: An increase in volume has opposite effects on the pressure of a gas. To evaluate the decrease in pressure, multiply the original pressure (101kPa) by the ratio of V₁ to V₂ (0.913) and the ratio of T₂ to T₁ (1.017). The result is 98.80 kPa
8. What is the new volume of a 15 L sample of gas at 365 K and 107 kPa when it is heated to 395 K and the pressure increases to 112 kPa?

Given: V₁= 15 L; T₁= 365 K; P₁= 107 kPa; T₂= 395 K; P₂= 112 kPa

Required: V₂

Solve: V₂= V₁ x P₁ x T₂ / P₂ x T₁

V₂ = 107 kPa x 15 L x 395 K / 365 K x 112 kPa = 15.51 L

Paraphrase: The new volume is 15.51 L

Evaluation: To evaluate the increase in volume, multiply the original volume (15L) by the ratio of P₁ to P₂ (0.95) and the ratio of T₂ to T₁ (1.082). The result is 15.51 L

9. What is the new temperature of a 4 L sample of gas at 375 K and 102 kPa when it is expanded to 9 L and has a pressure of 95 kPa?   

Given: V₁= 4 L; T₁= 375 K; P₁= 102 kPa; V₂= 9L; P₂= 95 kPa

Required: T₂

Solve: T₂= T₁ x P₂ x V₂ / V₁ x P₁

T₂ = 375 K x 95 kPa x 9 L / 4 L x 102 kPa = 785.9 K

Paraphrase: The new temperature is 785.9 k

Evaluation: To evaluate the increase in temperature, multiply the original temperature (375 K) by the ratio of P₂ to P₁ (0.93) and the ratio of V₂ to V₁ (2.25). The result is 785.9 K