Gases, solids, and liquids are the three states that a substance can exist as. A gas does not maintain shape and volume. Gases expand to fit and fill the container. While a liquid maintains volume, but not shape and fits the shape of the container it is placed in. Solids can maintain both volume and shape. When a gas exerts force on the walls of a container, this is known as its pressure. Pressure varies based on temperature and volume.
Many gas laws were used within this experiment, such as the Ideal Gas Law, Boyle’s Law, Charles’s Law, and Avogadro’s Law. The Ideal Gas Law is simply the combination of all the simple gas laws, Boyle’s Law, Charles’ Law, and Avogadro’s Law [which combines all of the properties of a gaseous sample].2 Boyle’s …show more content…
This is due to the volume in mL being multiplied by the pressure in kPa. Using the equation, PV = k, and the average constant, k, 1052.85 mL*kPa and the 2.5 mL volume, the pressure can be solved. The resulting pressure value is 421.14 kPa. When the volume is 40 mL, the pressure value is 26.32 kPa when the constant is the same value. The volume of 40 mL is 16 times as much as the volume of 2.5 mL. The pressure of the 40 mL was 1/16th of the pressure of the 2.5 mL. The pressure values of 2.5 and 40 mL were also 16 times greater than the other. When calculating the 5 mL volume it is twice the pressure value of 2.5 mL. Therefore, when the volume is 5 mL, the pressure reading is 210.57 kPa. When compared to the 2.5 mL value of 421.14 kPa, the 5 mL value is one half of the 2.5 mL pressure reading. For the 10 mL volume, the pressure is 105.29 kPa which is 1/4th the pressure of the 2.5 mL volume. The given volumes when multiplied by the calculated pressures equal the constant value of 1052.85 mL*kPa. These values are shown in Table A. This is seen because the pressure and volume are inversely proportional.
Table A: Pressure Values Calculated at Each Specific Volume, using Average Constant of 1052.85 mL*kPa
Volume, V (mL) Pressure, P (kPa) Constant, kPart A (mL*kPa)
2.5 421.14 1052.85
5 210.57 1052.85
10 105.29 1052.85
40 26.32
Many gas laws were used within this experiment, such as the Ideal Gas Law, Boyle’s Law, Charles’s Law, and Avogadro’s Law. The Ideal Gas Law is simply the combination of all the simple gas laws, Boyle’s Law, Charles’ Law, and Avogadro’s Law [which combines all of the properties of a gaseous sample].2 Boyle’s …show more content…
This is due to the volume in mL being multiplied by the pressure in kPa. Using the equation, PV = k, and the average constant, k, 1052.85 mL*kPa and the 2.5 mL volume, the pressure can be solved. The resulting pressure value is 421.14 kPa. When the volume is 40 mL, the pressure value is 26.32 kPa when the constant is the same value. The volume of 40 mL is 16 times as much as the volume of 2.5 mL. The pressure of the 40 mL was 1/16th of the pressure of the 2.5 mL. The pressure values of 2.5 and 40 mL were also 16 times greater than the other. When calculating the 5 mL volume it is twice the pressure value of 2.5 mL. Therefore, when the volume is 5 mL, the pressure reading is 210.57 kPa. When compared to the 2.5 mL value of 421.14 kPa, the 5 mL value is one half of the 2.5 mL pressure reading. For the 10 mL volume, the pressure is 105.29 kPa which is 1/4th the pressure of the 2.5 mL volume. The given volumes when multiplied by the calculated pressures equal the constant value of 1052.85 mL*kPa. These values are shown in Table A. This is seen because the pressure and volume are inversely proportional.
Table A: Pressure Values Calculated at Each Specific Volume, using Average Constant of 1052.85 mL*kPa
Volume, V (mL) Pressure, P (kPa) Constant, kPart A (mL*kPa)
2.5 421.14 1052.85
5 210.57 1052.85
10 105.29 1052.85
40 26.32