1. What pressure would be needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume?
2. If the pressure on a 1.04-L sample of gas is doubled at constant temperature, what will be the new volume of the gas?
3. A 1.04-L sample of gas at 759 mm Hg pressure is expanded until its volume is 2.24 L. What will be the pressure in the expanded gas sample (at constant temperature)?
4. What pressure would have to be applied to a 27.2-mL sample of gas at 25 degrees celsius and 1.00 atm to compress its volume to 1.00 mL without a change in temperature?
5. How would you describe absolute zero?
6. A favorite demonstration in introductory chemistry is to illustrate how the volume of a gas is affected by temperature by blowing up a balloon at room temperature and then placing the balloon into a container of dry ice or liquid nitrogen (both of which are very cold). Suppose a balloon containing 1.15 L of air at 25.2 degrees Celsius is placed into a flask containing liquid nitrogen at -78.5 degrees celsius. What will the volume of the sample become (at constant pressure)?
7.Suppose a 1.25 L of argon is cooled from 291 K to 78 K. What will be the new volume of the argon sample?
8. Suppose a gas sample is cooled from 600 K to 300 K. Ho will the new volume of the gas be related to its original volume?
9. Th label of an aerosol spray can contains a warning that the can should not be heated to over 130 degrees F because of the danger of explosion due to the pressure increase as it is heated. Calculate the potential volume of the gas contained in 500-mL aerosol can when it is heated from 25 degrees C to 54 degrees C (approx. 130 degrees F) assuming a constant pressure.
10. A sample of gas has a volume of 127 mL in a boiling water bath at 100 degrees celsius. Calculate the volume of the sample of gas 10 degrees celsius intervals after the heat source is turned off and the gas sample begins to cool down to eh temperature of the laboratory, 20 degrees celsius.
11. At conditions of constant temperature and pressure, the volume of a sample of ideal gas is _____ proportional to the number of moles of gas present.
12. A mathematical expression that summarizes Avogadro's law is ________.
13. If 0.105 mol of helium gas occupies a volume of 2.35 L at a certain temperature and pressure, what volume would 0.337 mol of helium occupy under the same conditions?
14. If 1.00 mol of helium occupies a volume of 22.4 L at 273 K at 1.00 atm, what volume will 1.00 g of helium will occupy under the same conditions?
15. If 3.25 mol of argon gas occupies a volume if 100. L at a particular and temperature and pressure, what volume does 14.15 mol of argon occupy under the same conditions?
16. If 2.71 g of argon gas occupies a volume of 4.21 L, what volume will 1.29 mol of argon occupy under the same conditions?
17. If gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in an evacuated 1.04-L container at 25 degrees celsius, what will be the partial pressure of each gas and the total pressure in the container?
18. A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of CO2 at 25 degrees celsius and a total pressure of 10.0 atm. Calculate the partial pressure (in torr) of each gas in the mixture.
19. How many moles of helium gas would be required 2.14-L container to a pressure of 759 mm Hg at 25 degrees celsius? How many moles of neon gas would be required to fill a similar tank to the same pressure at 25 degrees celsius?
20. Calculate what mass of argon gas is required to fill a 20.4-L container to a pressure of 1.09 atm at 25 degrees celsius.
21. What is the pressure in a 245-L tank that contains 5.21 kg of helium at 27 degrees celsius?
22. What mass of helium gas is needed to pressurize a 100.0-L tank to a 255 atm at 25 degrees celsius? What mass of oxygen gas would be needed to pressurize a similar tank to the same specifications?
23. At what temperature will a 1.0 g sample of neon gas exert a pressure of 500. torr in a 5.0 L container?
See the response by drwls at
http://www.jiskha.com/display.cgi?id=1268284727
1. To compress the volume of hydrogen from 25.1 mL to 25% of its original volume, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Since the volume is being compressed to 25% of its original volume, it will be reduced to 0.25 * 25.1 mL = 6.275 mL.
Using Boyle's Law:
P1 * V1 = P2 * V2
P1 = 1.01 atm (initial pressure)
V1 = 25.1 mL (initial volume)
V2 = 6.275 mL (final volume)
Substituting the values:
1.01 atm * 25.1 mL = P2 * 6.275 mL
Solving for P2:
P2 = (1.01 atm * 25.1 mL) / 6.275 mL
P2 = 4.06 atm
Therefore, the pressure needed to compress the volume of hydrogen to 25% of its original volume is approximately 4.06 atm.
2. According to the combined gas law, if the pressure on a gas is doubled at constant temperature, the new volume of the gas can be calculated using the equation:
P1 * V1 = P2 * V2
P1 = Initial pressure
V1 = Initial volume
P2 = New pressure (doubled)
V2 = New volume (to be determined)
Since the pressure is doubled, P2 = 2 * P1.
Substituting the values:
1.04 L * P1 = 2 * P1 * V2
P1 cancels out:
1.04 L = 2 * V2
Solving for V2:
V2 = 1.04 L / 2
V2 = 0.52 L
Therefore, the new volume of the gas will be 0.52 L.
3. To calculate the pressure in the expanded gas sample, we can use Boyle's Law again. Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature.
Using Boyle's Law:
P1 * V1 = P2 * V2
P1 = 759 mm Hg (initial pressure)
V1 = 1.04 L (initial volume)
V2 = 2.24 L (final volume)
Substituting the values:
759 mm Hg * 1.04 L = P2 * 2.24 L
Solving for P2:
P2 = (759 mm Hg * 1.04 L) / 2.24 L
P2 = 351 mm Hg
Therefore, the pressure in the expanded gas sample is approximately 351 mm Hg.
4. To calculate the pressure required to compress the volume of gas to 1.00 mL at constant temperature, we can use Boyle's Law.
Using Boyle's Law:
P1 * V1 = P2 * V2
P1 = 1.00 atm (initial pressure)
V1 = 27.2 mL (initial volume)
V2 = 1.00 mL (final volume)
Substituting the values:
1.00 atm * 27.2 mL = P2 * 1.00 mL
Solving for P2:
P2 = (1.00 atm * 27.2 mL) / 1.00 mL
P2 = 27.2 atm
Therefore, the pressure required to compress the volume of gas to 1.00 mL at constant temperature is approximately 27.2 atm.
5. Absolute zero refers to the lowest possible temperature that can be reached, at which a substance has minimal or no thermal energy. It is the point at which molecular motion virtually ceases. In terms of temperature, absolute zero is 0 Kelvin (K) or -273.15 degrees Celsius (°C) on the Kelvin and Celsius scales, respectively. At absolute zero, all substances theoretically reach their lowest possible energy state.
6. To calculate the new volume of the air sample when placed into a flask containing liquid nitrogen at -78.5 degrees Celsius, we can use Charles's Law. Charles's Law states that the volume and temperature of a gas are directly proportional at constant pressure.
Using Charles's Law:
V1 / T1 = V2 / T2
V1 = 1.15 L (initial volume)
T1 = 25.2 degrees Celsius + 273.15 (Kelvin conversion)
V2 = New volume (to be determined)
T2 = -78.5 degrees Celsius + 273.15 (Kelvin conversion)
Substituting the values:
1.15 L / (25.2 + 273.15) K = V2 / (-78.5 + 273.15) K
Simplifying the equation:
1.15 L / 298.35 K = V2 / 194.65 K
Cross-multiplying:
1.15 L * 194.65 K = V2 * 298.35 K
Solving for V2:
V2 = (1.15 L * 194.65 K) / 298.35 K
V2 = 0.748 L
Therefore, the volume of the air sample will become approximately 0.748 L when placed into the flask containing liquid nitrogen at -78.5 degrees Celsius.
7. To calculate the new volume of the argon sample when cooled from 291 K to 78 K, we can use Charles's Law again.
Using Charles's Law:
V1 / T1 = V2 / T2
V1 = 1.25 L (initial volume)
T1 = 291 K (initial temperature)
V2 = New volume (to be determined)
T2 = 78 K (new temperature)
Substituting the values:
1.25 L / 291 K = V2 / 78 K
Simplifying the equation:
1.25 L / 291 K = V2 / 78 K
Cross-multiplying:
1.25 L * 78 K = V2 * 291 K
Solving for V2:
V2 = (1.25 L * 78 K) / 291 K
V2 = 0.334 L
Therefore, the new volume of the argon sample will be approximately 0.334 L when cooled from 291 K to 78 K.
8. When a gas sample is cooled from 600 K to 300 K at constant pressure, its new volume will be directly proportional to its original volume. This is due to Charles's Law, which states that the volume of a gas is directly proportional to its temperature at constant pressure.
Therefore, the new volume of the gas will be equal to its original volume.