1. The volume of a gas is 10.0 liters and its pressure is 1.5 atm. If the pressure is decreased to 0.75 atm, what is its new volume?
2. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressure of the gas?
3. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg?
4. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?
5. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm.
6. The volume of a gas is 10.0 liters and its temperature is 1500 K. If the temperature is decreased to 750 K, what is its new volume?
7. The temperature of a gas is 100 K and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new temperature of the gas?
8. A gas occupies 12.3 liters at a temperature of 40.0 K. What is the volume when the temperature is increased to 60.0 K?
9. 500.0 mL of a gas is collected at 745 K. What will the volume be at standard temperature?
10. Convert 350.0 mL at 740 K to its new volume at standard temperature
2. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressure of the gas?
50kilopascal
1. To solve this problem, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law equation: P1V1 = P2V2
Given:
P1 = 1.5 atm
V1 = 10.0 L
P2 = 0.75 atm (new pressure)
Let's substitute the given values into the equation:
1.5 atm * 10.0 L = 0.75 atm * V2
15.0 L = 0.75 atm * V2
Now, isolate V2:
V2 = 15.0 L / 0.75 atm
V2 = 20.0 L
Therefore, the new volume of the gas is 20.0 liters.
2. To solve this problem, we can use the combined gas law, which states that the pressure, volume, and temperature of a gas are all related.
The combined gas law equation: P1V1 / T1 = P2V2 / T2
Given:
P1 = 100.0 kPa
V1 = 500.0 ml
V2 = 1000.0 ml (new volume)
Let's convert the volumes to liters:
V1 = 500.0 ml = 0.5 L
V2 = 1000.0 ml = 1.0 L
Substituting the given values into the equation:
(100.0 kPa * 0.5 L) / T1 = (P2 * 1.0 L) / T2
Now, let's solve for P2:
P2 = (100.0 kPa * 0.5 L * T2) / (T1 * 1.0 L)
P2 = (50.0 kPa * T2) / T1
Therefore, the new pressure of the gas is (50.0 kPa * T2) / T1.
3. To solve this problem, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law equation: P1V1 = P2V2
Given:
V1 = 12.3 L
P1 = 40.0 mm Hg
P2 = 60.0 mm Hg (new pressure)
Let's convert the pressures to atm:
P1 = 40.0 mm Hg = 40.0 mm Hg * 1 atm / 760 mm Hg = 0.0526 atm
P2 = 60.0 mm Hg = 60.0 mm Hg * 1 atm / 760 mm Hg = 0.0789 atm
Substituting the given values into the equation:
0.0526 atm * 12.3 L = 0.0789 atm * V2
Solving for V2:
V2 = (0.0526 atm * 12.3 L) / 0.0789 atm
V2 = 8.19 L
Therefore, the new volume of the gas is 8.19 liters.
4. To solve this problem, we can use the combined gas law, which states that the pressure, volume, and temperature of a gas are all related.
The combined gas law equation: P1V1 / T1 = P2V2 / T2
Given:
V1 = 3.60 L
P1 = 1.00 atm
P2 = 2.50 atm (new pressure)
Substituting the given values into the equation:
(1.00 atm * 3.60 L) / (25.0 °C) = (2.50 atm * V2) / (T2 °C)
Let's convert the temperatures to Kelvin:
T1 = 25.0 °C + 273.15 = 298.15 K
T2 = T2 °C + 273.15
Now, let's solve for V2:
(1.00 atm * 3.60 L) / 298.15 K = (2.50 atm * V2) / T2
V2 = (1.00 atm * 3.60 L * T2) / (2.50 atm * 298.15 K)
Therefore, the new volume of the gas is (1.00 atm * 3.60 L * T2) / (2.50 atm * 298.15 K).
5. To solve this problem, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law equation: P1V1 = P2V2
Given:
V1 = 1.56 L
P1 = 1.00 atm
P2 = 3.00 atm (new pressure)
Substituting the given values into the equation:
1.00 atm * 1.56 L = 3.00 atm * V2
Let's solve for V2:
V2 = (1.00 atm * 1.56 L) / 3.00 atm
Therefore, the new volume of the gas is (1.00 atm * 1.56 L) / 3.00 atm.
6. To solve this problem, we can use Charles's law, which states that the volume and temperature of a gas are directly proportional at constant pressure.
Charles's Law equation: V1 / T1 = V2 / T2
Given:
V1 = 10.0 L
T1 = 1500 K
T2 = 750 K (new temperature)
Substituting the given values into the equation:
10.0 L / 1500 K = V2 / 750 K
Let's solve for V2:
V2 = (10.0 L * 750 K) / 1500 K
Therefore, the new volume of the gas is (10.0 L * 750 K) / 1500 K.
7. To solve this problem, we can use Charles's law, which states that the volume and temperature of a gas are directly proportional at constant pressure.
Charles's Law equation: V1 / T1 = V2 / T2
Given:
T1 = 100 K
V1 = 500.0 ml
V2 = 1000.0 ml (new volume)
Let's convert the volumes to liters:
V1 = 500.0 ml = 0.5 L
V2 = 1000.0 ml = 1.0 L
Substituting the given values into the equation:
0.5 L / (100 K) = 1.0 L / T2
Now, let's solve for T2:
T2 = (1.0 L * 100 K) / 0.5 L
Therefore, the new temperature of the gas is (1.0 L * 100 K) / 0.5 L.
8. To solve this problem, we can use Charles's law, which states that the volume and temperature of a gas are directly proportional at constant pressure.
Charles's Law equation: V1 / T1 = V2 / T2
Given:
V1 = 12.3 L
T1 = 40.0 K
T2 = 60.0 K (new temperature)
Substituting the given values into the equation:
12.3 L / 40.0 K = V2 / 60.0 K
Let's solve for V2:
V2 = (12.3 L * 60.0 K) / 40.0 K
Therefore, the new volume of the gas is (12.3 L * 60.0 K) / 40.0 K.
9. To solve this problem, we can use Gay-Lussac's law, which states that the pressure and temperature of a gas are directly proportional at constant volume.
Gay-Lussac's Law equation: P1 / T1 = P2 / T2
Given:
V1 = 500.0 mL
T1 = 745 K
T2 = standard temperature (273.15 K)
Let's convert the volume to liters:
V1 = 500.0 mL = 0.5 L
Substituting the given values into the equation:
P1 / 745 K = P2 / 273.15 K
Now, let's solve for P2:
P2 = (P1 * 273.15 K) / 745 K
Therefore, the new pressure of the gas at standard temperature is (P1 * 273.15 K) / 745 K.
10. To solve this problem, we can use Gay-Lussac's law, which states that the pressure and temperature of a gas are directly proportional at constant volume.
Gay-Lussac's Law equation: P1 / T1 = P2 / T2
Given:
V1 = 350.0 mL
T1 = 740 K
T2 = standard temperature (273.15 K)
Let's convert the volume to liters:
V1 = 350.0 mL = 0.35 L
Substituting the given values into the equation:
P1 / 740 K = P2 / 273.15 K
Now, let's solve for P2:
P2 = (P1 * 273.15 K) / 740 K
Therefore, the new pressure of the gas at standard temperature is (P1 * 273.15 K) / 740 K.
1. To find the new volume when the pressure is decreased, you can use Boyle's Law, which states that the product of the initial volume and pressure is equal to the product of the final volume and pressure.
In this case, the initial volume (V1) is 10.0 liters and the initial pressure (P1) is 1.5 atm. The final pressure (P2) is 0.75 atm, and we need to find the final volume (V2).
Using Boyle's Law equation: P1 * V1 = P2 * V2
Solving for V2: V2 = (P1 * V1) / P2
Substituting the given values: V2 = (1.5 atm * 10.0 L) / 0.75 atm
V2 = 20.0 liters
Therefore, the new volume is 20.0 liters.
2. To find the new pressure when the volume increases, you can use Boyle's Law again.
In this case, the initial pressure (P1) is 100.0 kPa and the initial volume (V1) is 500.0 ml. The final volume (V2) is 1,000.0 ml, and we need to find the final pressure (P2).
Using Boyle's Law equation: P1 * V1 = P2 * V2
Solving for P2: P2 = (P1 * V1) / V2
Substituting the given values: P2 = (100.0 kPa * 500.0 ml) / 1,000.0 ml
P2 = 50.0 kPa
Therefore, the new pressure is 50.0 kPa.
3. To find the new volume when the pressure is increased, you can use Boyle's Law once again.
In this case, the initial volume (V1) is 12.3 liters and the initial pressure (P1) is 40.0 mm Hg. The final pressure (P2) is 60.0 mm Hg, and we need to find the final volume (V2).
Using Boyle's Law equation: P1 * V1 = P2 * V2
Solving for V2: V2 = (P1 * V1) / P2
Substituting the given values: V2 = (40.0 mm Hg * 12.3 L) / 60.0 mm Hg
V2 = 8.20 liters
Therefore, the new volume is 8.20 liters.
4. To find the new volume when the pressure changes, you can use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law.
In this case, the initial volume (V1) is 3.60 liters and the initial pressure (P1) is 1.00 atm. The final pressure (P2) is 2.50 atm, and we need to find the final volume (V2).
Using the combined gas law equation: (P1 * V1) / T1 = (P2 * V2) / T2
Since the temperature is not given, we can assume it remains constant.
Simplifying the equation: (P1 * V1) = (P2 * V2)
Solving for V2: V2 = (P1 * V1) / P2
Substituting the given values: V2 = (1.00 atm * 3.60 L) / 2.50 atm
V2 = 1.44 liters
Therefore, the new volume is 1.44 liters.
5. To find the new volume when the pressure changes, you can once again use the combined gas law.
In this case, the initial volume (V1) is 1.56 L and the initial pressure (P1) is 1.00 atm. The final pressure (P2) is 3.00 atm, and we need to find the final volume (V2).
Using the combined gas law equation: (P1 * V1) / T1 = (P2 * V2) / T2
Since the temperature is not given, we can assume it remains constant.
Simplifying the equation: (P1 * V1) = (P2 * V2)
Solving for V2: V2 = (P1 * V1) / P2
Substituting the given values: V2 = (1.00 atm * 1.56 L) / 3.00 atm
V2 = 0.52 liters
Therefore, the new volume is 0.52 liters.
6. To find the new volume when the temperature is decreased, you can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature.
In this case, the initial volume (V1) is 10.0 liters and the initial temperature (T1) is 1500 K. The final temperature (T2) is 750 K, and we need to find the final volume (V2).
Using Charles's Law equation: (V1 / T1) = (V2 / T2)
Solving for V2: V2 = (V1 * T2) / T1
Substituting the given values: V2 = (10.0 L * 750 K) / 1500 K
V2 = 5.00 liters
Therefore, the new volume is 5.00 liters.
7. To find the new temperature when the volume increases, you can use Charles's Law once again.
In this case, the initial temperature (T1) is 100 K and the initial volume (V1) is 500.0 ml. The final volume (V2) is 1,000.0 ml, and we need to find the final temperature (T2).
Using Charles's Law equation: (V1 / T1) = (V2 / T2)
Solving for T2: T2 = (V2 * T1) / V1
Substituting the given values: T2 = (1,000.0 ml * 100 K) / 500.0 ml
T2 = 200 K
Therefore, the new temperature is 200 K.
8. To find the new volume when the temperature is increased, you can again use Charles's Law.
In this case, the initial volume (V1) is 12.3 liters and the initial temperature (T1) is 40.0 K. The final temperature (T2) is 60.0 K, and we need to find the final volume (V2).
Using Charles's Law equation: (V1 / T1) = (V2 / T2)
Solving for V2: V2 = (V1 * T2) / T1
Substituting the given values: V2 = (12.3 L * 60.0 K) / 40.0 K
V2 = 18.45 liters
Therefore, the new volume is 18.45 liters.
9. To find the volume at standard temperature (0°C or 273.15 K), you can use Gay-Lussac's Law, which states that the pressure and volume of a gas are directly proportional at constant temperature.
In this case, the initial volume (V1) is 500.0 mL and the initial temperature (T1) is 745 K. We need to find the final volume (V2) at standard temperature (T2 = 273.15 K).
Using Gay-Lussac's Law equation: (V1 / T1) = (V2 / T2)
Solving for V2: V2 = (V1 * T2) / T1
Substituting the given values: V2 = (500.0 mL * 273.15 K) / 745 K
V2 ≈ 183.29 mL
Therefore, the volume at standard temperature is approximately 183.29 mL.
10. To convert the volume from a given temperature to the volume at standard temperature, you can again use Gay-Lussac's Law.
In this case, the initial volume (V1) is 350.0 mL and the initial temperature (T1) is 740 K. We need to find the final volume (V2) at standard temperature (T2 = 273.15 K).
Using Gay-Lussac's Law equation: (V1 / T1) = (V2 / T2)
Solving for V2: V2 = (V1 * T2) / T1
Substituting the given values: V2 = (350.0 mL * 273.15 K) / 740 K
V2 ≈ 128.35 mL
Therefore, the new volume at standard temperature is approximately 128.35 mL.