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 (P1*V1)/T1 = (P2*V2)/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 = 13°C + 273.15 = 286.15 K
T2 = 25.7°C + 273.15 = 298.85 K

Step 2: Convert the pressure of the gas to kPa. The problem states that the gas is collected over water, so the pressure of the gas is the atmospheric pressure minus the vapor pressure of water.

P1 = 101.4 kPa - 9.1 kPa = 92.3 kPa
P2 = 103 kPa (already in kPa, no conversion needed)

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

(92.3 kPa * 5.6 L) / 286.15 K = (103 kPa * V2) / 298.85 K

Solving for V2 gives:

V2 = (92.3 kPa * 5.6 L * 298.85 K) / (103 kPa * 286.15 K) = 5.37 L

So, the volume of the dried gas at 25.7 °C and 103 kPa is approximately 5.37 L.

Question

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 (P1*V1)/T1 = (P2*V2)/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 = 13°C + 273.15 = 286.15 K
T2 = 25.7°C + 273.15 = 298.85 K

Step 2: Convert the pressure of the gas to kPa. The problem states that the gas is collected over water, so the pressure of the gas is the atmospheric pressure minus the vapor pressure of water.

P1 = 101.4 kPa - 9.1 kPa = 92.3 kPa
P2 = 103 kPa (already in kPa, no conversion needed)

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

(92.3 kPa * 5.6 L) / 286.15 K = (103 kPa * V2) / 298.85 K

Solving for V2 gives:

V2 = (92.3 kPa * 5.6 L * 298.85 K) / (103 kPa * 286.15 K) = 5.37 L

So, the volume of the dried gas at 25.7 °C and 103 kPa is approximately 5.37 L.

Knowee AI · Accepted Answer

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 (P1*V1)/T1 = (P2*V2)/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 = 13°C + 273.15 = 286.15 K
T2 = 25.7°C + 273.15 = 298.85 K

Step 2: Convert the pressure of the gas to kPa. The problem states that the gas is collected over water, so the pressure of the gas is the atmospheric pressure minus the vapor pressure of water.

P1 = 101.4 kPa - 9.1 kPa = 92.3 kPa
P2 = 103 kPa (already in kPa, no conversion needed)

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

(92.3 kPa * 5.6 L) / 286.15 K = (103 kPa * V2) / 298.85 K

Solving for V2 gives:

V2 = (92.3 kPa * 5.6 L * 298.85 K) / (103 kPa * 286.15 K) = 5.37 L

So, the volume of the dried gas at 25.7 °C and 103 kPa is approximately 5.37 L.

If 5.6 L of a gas is collected over water at 13°C at an atmospheric pressure of 101.4 kPa, what is the volume of the dried gas at 25.7 °C and 103 kPa? (The vapour pressure of water is 9.1 kPa.)

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