Gas Laws Quiz

Oxygen gas is at a pressure of 4.0 atm when it occupies a volume of 5.5 liters. What will be the pressure when the gas expands to a volume of 9.0 liters?

• 2.44 atm

• 3.00 atm

• 4.50 atm

• 5.00 atm

Answer explanation

According to Boyle's Law, P1V1 = P2V2. Solving for P2, we get P2 = (P1V1)/V2 = (4.0 atm * 5.5 L)/9.0 L = 2.44 atm. Therefore, the pressure will be 2.44 atm when the gas expands to a volume of 9.0 liters.

Match the following volumes with the corresponding pressures.

• A sample of nitrogen occupies a volume of 125 L at what pressure if it was 125 L at 250 torr?

• 62.5 L

• A sample of nitrogen occupies a volume of 500 L at what pressure if it was 125 L at 250 torr?

• 500 L

• A sample of nitrogen occupies a volume of 62.5 L at what pressure if it was 125 L at 250 torr?

• 125 L

• A sample of nitrogen occupies a volume of 250 L at what pressure if it was 125 L at 250 torr?

• 250 L

Answer explanation

According to Boyle's Law, P1V1 = P2V2. Therefore, (250 torr)(125 L) = (500 torr)(V2). Solving for V2 gives V2 = 62.5 L.

Oxygen gas is at a pressure of 15 atm when it occupies a volume of 2.5 L. To what pressure should it be changed to in order to occupy a volume of 9.9 L?

• 3.79 atm

• 4.5 atm

• 5.2 atm

• 6.1 atm

Answer explanation

To find the new pressure, we use Boyle's Law: P1V1 = P2V2. Plugging in the values, we get (15 atm)(2.5 L) = (P2)(9.9 L). Solving for P2 gives P2 = 3.79 atm.

Chlorine gas occupies a volume of 25 mL at 350 K. What volume in mL will it occupy at 690 K?

• 50 mL

• 35 mL

• 40 mL

• 45 mL

Answer explanation

According to Charles's Law, V1/T1 = V2/T2. Therefore, (25 mL)/(350 K) = V2/(690 K). Solving for V2 gives V2 = 50 mL.

A sample of neon gas at 45°C and a volume of 3.5 L is cooled to 15°C. What is the new volume?

• 2.5 L

• 3.0 L

• 3.5 L

• 4.0 L

Answer explanation

When a gas is cooled at constant pressure, the volume is directly proportional to the temperature. Therefore, the new volume will remain the same at 3.5 L.

Fluorine gas at 699 K occupies a volume of 455 mL. To what temperature should it be lowered to bring the volume to 255 mL?

• 155 K

• 312 K

• 391 K

• 475 K

Answer explanation

To find the new temperature, we use the combined gas law formula (P1V1/T1) = (P2V2/T2). Solving for T2, we get T2 = (P2V2*T1)/(P1V1) = (1*255*699)/(1*455) = 391 K.

Helium occupies a volume of 3.8 liters at -45°C. What volume will it occupy at 140°C?

• 2.5 liters

• 4.8 liters

• 5.3 liters

• 6.1 liters

Answer explanation

As temperature increases, volume also increases. Using Charles's Law, V1/T1 = V2/T2, we find V2 = (V1 * T2) / T1 = (3.8 * 413K) / 228K = 4.8 liters.

Match the following temperatures with the conditions under which the aerosol can will burst.

• Temperature at which the aerosol can remains stable under initial conditions.

• 298 Kelvin

• Temperature not related to the bursting condition of the aerosol can.

• 350 Kelvin

• Temperature not related to the bursting condition of the aerosol can.

• 596 Kelvin

• Temperature at which the aerosol can will burst if the pressure exceeds 30 atmospheres.

• 450 Kelvin

Answer explanation

The aerosol can will burst if the pressure exceeds 30 atmospheres. Using the ideal gas law, we find that the temperature required for the can to burst is 596 Kelvin.

A gas cylinder in a dentist's office contains dinitrogen oxide gas at 5.0 atmospheres at 20.0°C. If the dentist office catches fire and the gas cylinder is surrounded by 1400°C flames, the pressure inside the cylinder will be (a)   .

• Significantly higher than 5.0 atmos
• Significantly lower than 5.0 atmosp
• Remains the same at 5.0 atmospheres
• Cannot be determined without additi

Answer explanation

The pressure inside the gas cylinder will significantly increase due to the high temperature, causing the gas to expand and the pressure to rise above 5.0 atmospheres.

A playground ball is filled to 760 mmHg with nitrogen at 17°C. The temperature in Kelvin required to increase the pressure inside the ball to 1000 mmHg is (a)   .

• 223 K
• 273 K
• 298 K
• 323 K

Answer explanation

To increase pressure, use Gay-Lussac's Law: P1/T1 = P2/T2. Convert temperatures to Kelvin. (760/290) = (1000/T2). Solve for T2 to get 298 K.