In an experiment, 25.0 mL of a gas with a pressure of 1.00 atm is contained in a balloon at 25.0°C. The balloon is then cooled to 5.0°C, and the pressure is found to be 0.750 atm. What is the volume of the gas under the new conditions?
To find the volume of the gas under the new conditions, we can use the combined gas law equation, which relates the initial and final conditions of pressure, temperature, and volume:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where P1, V1, and T1 represent the initial conditions, and P2, V2, and T2 represent the final conditions.
Given:
P1 = 1.00 atm (Initial pressure)
V1 = 25.0 mL (Initial volume)
T1 = 25.0°C (Initial temperature)
P2 = 0.750 atm (Final pressure)
T2 = 5.0°C (Final temperature)
First, we need to convert the temperatures to Kelvin scale by adding 273.15 to each temperature:
T1 = 25.0°C + 273.15 = 298.15 K
T2 = 5.0°C + 273.15 = 278.15 K
Now, we can plug these values into the combined gas law equation and solve for V2:
(1.00 atm * 25.0 mL) / (298.15 K) = (0.750 atm * V2) / (278.15 K)
To solve for V2, we can rearrange the equation and cross-multiply:
(1.00 atm * 25.0 mL * 278.15 K) = (0.750 atm * V2 * 298.15 K)
Divide both sides of the equation by (0.750 atm * 298.15 K):
V2 = (1.00 atm * 25.0 mL * 278.15 K) / (0.750 atm * 298.15 K)
Now, we can calculate V2:
V2 = (1.00 * 25.0 * 278.15) / (0.750 * 298.15)
= 9.34 mL
Therefore, the volume of the gas under the new conditions is approximately 9.34 mL.
(P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin.