Chemistry/Physics Question involving stopcocks which i dont understand?

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gp4rts

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Do you mean it is heated to 400 ºK?

The condition in the flask after the heating is

T 1= 400 ºK
P1 = 1 atm
V1 = 2 L

After cooling to 300 ºK

T2 = 300 ºK
P2 = ?
V2 = 2 L

The gas law states that P*V/T = constant so P1*V1/T1 = P2*V2/T2

P2 = P1*V1/T1 * T2/V2. Since V1 = V2 P2 = P1*T2/T1

P2 = 0.75 atm

The no of moles of oxygen is, from P*V = n*R*T

n = P*V/(R*T)

for R use 0.082 L*atm/ºK-mol

n = 0.75*2/(0.082*300) = 0.061 moles

The molecular mass of O2 is 32 g/mole so the mass is 0.061*32 = 1.95 g
 
A flask of volume 2L (2x10^-3 m^3) provided with a stopcock contains oxygen at 300K and athmospheric pressure. The system is heated to a temp of 40K, with the stopcock open to the atmosphere. The stopcock is then closed and the flask cooled to its original temperature.
A)WHat is the final pressure of the Oxygen in the flask?
B)How many grams of oxygen remains in the flask?
 
Do you mean it is heated to 400 ºK?

The condition in the flask after the heating is

T 1= 400 ºK
P1 = 1 atm
V1 = 2 L

After cooling to 300 ºK

T2 = 300 ºK
P2 = ?
V2 = 2 L

The gas law states that P*V/T = constant so P1*V1/T1 = P2*V2/T2

P2 = P1*V1/T1 * T2/V2. Since V1 = V2 P2 = P1*T2/T1

P2 = 0.75 atm

The no of moles of oxygen is, from P*V = n*R*T

n = P*V/(R*T)

for R use 0.082 L*atm/ºK-mol

n = 0.75*2/(0.082*300) = 0.061 moles

The molecular mass of O2 is 32 g/mole so the mass is 0.061*32 = 1.95 g
 
Do you mean it is heated to 400 ºK?

The condition in the flask after the heating is

T 1= 400 ºK
P1 = 1 atm
V1 = 2 L

After cooling to 300 ºK

T2 = 300 ºK
P2 = ?
V2 = 2 L

The gas law states that P*V/T = constant so P1*V1/T1 = P2*V2/T2

P2 = P1*V1/T1 * T2/V2. Since V1 = V2 P2 = P1*T2/T1

P2 = 0.75 atm

The no of moles of oxygen is, from P*V = n*R*T

n = P*V/(R*T)

for R use 0.082 L*atm/ºK-mol

n = 0.75*2/(0.082*300) = 0.061 moles

The molecular mass of O2 is 32 g/mole so the mass is 0.061*32 = 1.95 g
 
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