Exercises
Exercise 1. A gas has a pressure of 675 mm Hg in a volume of 455 mL. What is the pressure of the gas if the volume is changed to 250. mL? Assume a constant temperature. Which law does this follow?
\(\displaystyle P_1V_1\;=\;P_2V_2\)
P1 = 675 mm Hg, V1 = 455 mL = 0.455 L, V2 = 250. mL = 0.250 L, P2 = ??
Solve the equation for P2
\(\displaystyle P_2\;=\;\frac{P_1V_1}{V_2}\;=\frac{675\;mm\;Hg\times\;0.455L}{0.250\;L}\;=\;\mathbf{1.23\times\;10^3\;mm\;Hg}\)
This is Boyles Law.
Exercise 2. If gas at 75.0°C has a volume of 255 mL. The temperature is increased to 125.2°C, what is the volume? Assume the pressure is constant. Which law does this follow?
This is Charles’s Law.
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\)
V1 = 0.255 L, T1 = 75.0°C + 273.15 = 348.2 K, V2 = ??, T2 = 398.4 K.
Solve the above equation for V2.
\(\displaystyle V_2\;=\;\frac{V_1T_2}{T_1}\;=\frac{0.255\;L\times\;398.4\;K}{348.2\;K}\;=\;\mathbf{0.292\;L}\)Exercise 3. One mole of gas at STP occupies a volume of 22.41 L. What is the volume at 25.0°C and 760 mm Hg?
\(\displaystyle \frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)
P1 = 1.00 atm, V1 = 22.41 L, T1 = 273.15 K, P2 = 760 mm Hg = 1.00 atm, V2 = ???, T2 = 298.15 K
The pressure is the same on both sides, therefore our equation is:
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\)
Solve for V2
\(\displaystyle V_2\;=\;\frac{V_1T_2}{T_1}\;=\;\frac{22.41\;L\;\times\;298.15\;K}{273.15\;K}\;=\;\mathbf{22.46\;L}\)
Exercise 4. A sample of nitrogen gas at 19.0°C and 760 mm Hg has a volume of 2.83 mL. What is the volume at STP?
The amount of gas and the pressure is constant because 760 mm Hg is equal to 1.00 atm.
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\)
V1 = 2.83 mL = 0.00283 L, T1 = 19.0°C + 273.15 K = 292.15 K, V2 = ??, T2 = 273.15 K
Solve the equation for V2
\(\displaystyle V_2\;=\;\frac{V_1T_2}{T_1}\;=\;\frac{0.00283\;L\;\times\;273.15\;K}{292.15\;K}\;=\;\mathbf{0.00265\;L}\)
Exercise 5. A sample of a B vitamin, pantothenic acid, has a mass of 72.52 mg and gives 3.87 mL of nitrogen gas at 22.7°C at 784 mm Hg. What is the volume of nitrogen at STP?
The amount of gas and the pressure is constant.
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\)
V1 = 3.87 mL = 0.00387 L, T1 = 22.7°C + 273.15 K = 295.85 K, V2 = ??, T2 = 273.15 K
Solve the equation for V2
\(\displaystyle V_2\;=\;\frac{V_1T_2}{T_1}\;=\;\frac{0.00387\;L\;\times\;273.15\;K}{295.85\;K}\;=\;\mathbf{0.00357\;L}\)
Back to Gas Laws
Exercise 6. A sample of helium has a volume of 475 mL at 45.3°C and 742 mm Hg. The temperature is decreased to 21.0°C and the volume is increased to 555 mL. What is the new pressure?
\(\displaystyle \frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)
P1 = 742 mm Hg, V1 = 475 mL = 0.475 L, T1 = 45.3°C + 273.15 = 318.45 K, P2 = ??, V2 = 555 mL = 0.555 L, T2 = 21.0°C + 273.15 = 294.15 K
Solve the equation for P2
\(\displaystyle P_2\;=\;\frac{P_1V_1T_2}{T_1V_2}\;=\;\frac{742\;mm\;Hg\times\;0.475\;L\times\;294.15\;K}{318.45\;K\times\;0.555\;L}\;=\;\mathbf{586\;mm\;Hg}\)
Exercise 7. An experiment called for 5.75 L of sulfur dioxide, SO2, at 0.00°C and 1.00 atm of pressure. What is the volume at 25.00°C and 3.62 atm?
The amount of gas is constant.
\(\displaystyle \frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)
P1 = 1.00 atm, V1 = 5.75 L, T1 = 0.00°C + 273.15 = 273.15 K, P2 = 3.62 atm, V2 = ????, T2 = 25.00°C + 273.15 = 298.15 K
\(\displaystyle V_2\;=\;\frac{P_1V_1T_2}{T_1P_2}\;=\;\frac{1.00\;atm\times\;5.75\;L\times\;298.15\;K}{273.15\;K\times\;3.62\;atm}\;=\;\mathbf{1.73\;L}\)
Back to Gas Laws
Exercise 8. A hot air balloon has a volume of 136 L on a 22.0°C day. What would the volume be if the temperature plummeted to -3.6°C?
The amount of gas and the pressure is constant.
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\)
V1 = 136 L, T1 = 22.0°C + 273.15 K = 295.15 K, V2 = ??, T2 = -3.6°C + 273.15 = 269.5 K
Solve the equation for V2
\(\displaystyle V_2\;=\;\frac{V_1T_2}{T_1}\;=\;\frac{136\;L\;\times\;269.5\;K}{295.15\;K}\;=\;\mathbf{124\;L}\)
Exercise 9. The pressure of a gas is reduced from 1205.0 mm Hg to 795.0 mm Hg as the volume is increased by the movement of a piston from 83.5 to 365.0 mL. The original temperature was 86.2°C. What is the final temperature?
The amount of gas is constant.
\(\displaystyle \frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)
P1 = 1205.0 mm Hg, V1 = 83.5 mL = 0.0835 L, T1 = 86.2°C + 273.15 = 359.4 K, P2 = 795 mm Hg, V2 = 365 mL = 0.365 L, T2 = ???
Solve for T2
\(\displaystyle T_2\;=\;\frac{P_2V_2T_1}{P_1V_1}\;=\;\frac{795 mm Hg\;atm\times\;0.365\;L\times\;359.4\;K}{1205.0\;mm\;Hg\times\;0.0835\;L}\;=\;1036.5\;K\)
1036.5 K – 273.15 = 763°C
Exercise 10. A scuba tank with a volume of 13.0 L has a pressure of 156 atm. If the water temperature is 28.0°C, how many liters of air will the tank provide to the diver’s lungs at a depth of 75 feet in the ocean where the pressure is 3.25 atm?
The amount of gas is constant.
\(\displaystyle \frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)
P1 = 156 atm, V1 = 13.0 L, T1 = 28.0°C + 273.15 = 296.15 K, P2 = 3.25 atm, V2 = ????, T2 = 37.0°C + 273.15 = 310.15 K
\(\displaystyle V_2\;=\;\frac{P_1V_1T_2}{T_1P_2}\;=\;\frac{156\;atm\times\;13.0\;L\times\;310.15\;K}{296.15\;K\times\;3.25\;atm}\;=\;\mathbf{653\;L}\)