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 these types of gas law problems.
T1 = 21.3°C + 273.15 = 294.45 K
T2 = 25.3°C + 273.15 = 298.45 K

Step 2: Convert all pressures to the same units. In this case, we will use kPa.
P1 = 109.7 kPa - 1.7 kPa (the vapour pressure of water) = 108 kPa
P2 = 101.3 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2.
(V1*P1)/T1 = (V2*P2)/T2
(437 L * 108 kPa) / 294.45 K = (V2 * 101.3 kPa) / 298.45 K
V2 = (437 L * 108 kPa * 298.45 K) / (294.45 K * 101.3 kPa)
V2 = 450 L

So, the volume of the dried gas at 25.3 °C and 101.3 kPa is 450 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 these types of gas law problems. 
T1 = 21.3°C + 273.15 = 294.45 K
T2 = 25.3°C + 273.15 = 298.45 K

Step 2: Convert all pressures to the same units. In this case, we will use kPa. 
P1 = 109.7 kPa - 1.7 kPa (the vapour pressure of water) = 108 kPa
P2 = 101.3 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2.
(V1*P1)/T1 = (V2*P2)/T2
(437 L * 108 kPa) / 294.45 K = (V2 * 101.3 kPa) / 298.45 K
V2 = (437 L * 108 kPa * 298.45 K) / (294.45 K * 101.3 kPa)
V2 = 450 L

So, the volume of the dried gas at 25.3 °C and 101.3 kPa is 450 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 these types of gas law problems. 
T1 = 21.3°C + 273.15 = 294.45 K
T2 = 25.3°C + 273.15 = 298.45 K

Step 2: Convert all pressures to the same units. In this case, we will use kPa. 
P1 = 109.7 kPa - 1.7 kPa (the vapour pressure of water) = 108 kPa
P2 = 101.3 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2.
(V1*P1)/T1 = (V2*P2)/T2
(437 L * 108 kPa) / 294.45 K = (V2 * 101.3 kPa) / 298.45 K
V2 = (437 L * 108 kPa * 298.45 K) / (294.45 K * 101.3 kPa)
V2 = 450 L

So, the volume of the dried gas at 25.3 °C and 101.3 kPa is 450 L.

If 437 L of a gas is collected over water at 21.3°C at an atmospheric pressure of 109.7 kPa, what is the volume of the dried gas at 25.3 °C and 101.3 kPa? (The vapour pressure of water is 1.7 kPa.)

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