If 2.1 L of a gas is collected over water at 29.3°C at an atmospheric pressure of 103.2 kPa, what is the volume of the dried gas at 25.4 °C and 100.1 kPa? (The vapour pressure of water is 1 kPa.)

To solve this problem, we will use the combined gas law which states that the ratio of the product of pressure and volume to temperature of one state is equal to the ratio of the product of pressure and volume to temperature of another state. The formula is (P1V1)/T1 = (P2V2)/T2.

Step 1: Convert all temperatures to Kelvin. The Kelvin scale is used in gas law problems. To convert from Celsius to Kelvin, add 273.15.

T1 = 29.3°C + 273.15 = 302.45 K T2 = 25.4°C + 273.15 = 298.55 K

Step 2: Adjust the initial pressure to account for the vapour pressure of water. The total pressure is the sum of the pressure of the gas and the pressure of the water vapor. We need to find the pressure of the gas alone.

P1 (gas) = P1 (total) - P (water) P1 (gas) = 103.2 kPa - 1 kPa = 102.2 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2.

(102.2 kPa * 2.1 L) / 302.45 K = (100.1 kPa * V2) / 298.55 K

Cross multiply and solve for V2:

V2 = (102.2 kPa * 2.1 L * 298.55 K) / (302.45 K * 100.1 kPa)

V2 = 2.2 L

So, the volume of the dried gas at 25.4 °C and 100.1 kPa is 2.2 L.