-A pressurized can of whipping cream has an internal pressure of 1.087 atm at 23°C. If it is placed in a freezer at -14°C, what is the new value for its internal pressure?
-Carbon monoxide and molecular oxygen react to form carbon dioxide. A 54.0 L reactor at 200°C is charged with 1.04 atm of CO. The gas is then pressurized with O2 to give a total pressure of 3.52 atm. The reactor is sealed, heated to 350°C to drive the reaction to completion, and cooled back to 200°C. Compute the final partial pressure of each gas.
-Carbon monoxide and molecular oxygen react to form carbon dioxide. A 54.0 L reactor at 200°C is charged with 1.04 atm of CO. The gas is then pressurized with O2 to give a total pressure of 3.52 atm. The reactor is sealed, heated to 350°C to drive the reaction to completion, and cooled back to 200°C. Compute the final partial pressure of each gas.
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I'll help you with the first part.
-A pressurized can of whipping cream has an internal pressure of 1.087 atm at 23°C. If it is placed in a freezer at -14°C, what is the new value for its internal pressure?
PV = nRT
P1/T1 = (nR)/V1
P2/T2 = (nR)/V2
At constant volume, (nR)/V1 = (nR)/V2, so we can set the two expressions equal to each other:
P1/T1 = P2/T2
P1 = 1.087 atm
T1 = 23 °C = (23 + 273.15) = 296.15 K
P2 = unknown
T2 = -14 °C = (-14 + 273.15) = 259.15 K
P2 = (T2)*(P1/T1)
Now just plug in the numbers and calculate the answer.
If I have some time, I'll come back and help on the second one.
-A pressurized can of whipping cream has an internal pressure of 1.087 atm at 23°C. If it is placed in a freezer at -14°C, what is the new value for its internal pressure?
PV = nRT
P1/T1 = (nR)/V1
P2/T2 = (nR)/V2
At constant volume, (nR)/V1 = (nR)/V2, so we can set the two expressions equal to each other:
P1/T1 = P2/T2
P1 = 1.087 atm
T1 = 23 °C = (23 + 273.15) = 296.15 K
P2 = unknown
T2 = -14 °C = (-14 + 273.15) = 259.15 K
P2 = (T2)*(P1/T1)
Now just plug in the numbers and calculate the answer.
If I have some time, I'll come back and help on the second one.