1 L FLASK CONTAINS NITROGEN ALONG WITH A DROP OR TWO OF WATER AT 40C THE TOTAL PRESSURE BEING 760 TORR .IF ALL THE CONTENTS ARE TRANSFERRED TO ANOTHER FLASK OF 0.5L AT THE SAME TEMPERATURE WHAT WILL BE THE PRESSURE SET UP IN THE SECOND FLASK?
(AQUEOUS TENSION AT 40C = 55 TORR)
Well, let's break this down, shall we? First of all, we have a flask with nitrogen and a drop or two of water. A flask full of party-loving pants! Now, the total pressure in the first flask is 760 Torr. Quite impressive, I must say.
Now, we need to transfer all these party animals to another flask, which is half the size of the original flask. A cozy little party, you might say. Both flasks are at the same temperature, 40°C (aka, the perfect temperature for a hot tub party).
Now, here's where things get interesting. You see, water can be a real diva and creates what we call "aqueous tension" at high temperatures, which is around 55 Torr in this case. So, we can say that the water in the first flask is causing a bit of a "pressure drop".
When we transfer everything to the second flask, the total party pressure will be affected by that aqueous tension. So, we subtract the aqueous tension from the total pressure in the first flask (760 Torr - 55 Torr) to get the pressure in the second flask.
Therefore, the pressure in the second flask will be 705 Torr. So, get ready for a slightly less wild party, but still enough pressure to have a good time!
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L * atm / mol * K)
T is the temperature in Kelvin
Let's calculate the number of moles of nitrogen in the initial flask:
Given:
Initial volume (V1) = 1 L
Total pressure (P1) = 760 torr
Temperature (T) = 40°C = 313.15 K
To convert torr to atm, we divide by 760:
P1 = 760 torr / 760 = 1 atm
Using the ideal gas law, rearranged to solve for moles (n):
n = PV / RT
n1 = (P1 * V1) / (R * T)
Substituting the given values:
n1 = (1 atm * 1 L) / (0.0821 L * atm / mol * K * 313.15 K)
Calculating n1:
n1 = 0.0396 mol
Now, let's calculate the number of moles of water in the initial flask:
Given:
Aqueous tension at 40°C = 55 torr
Using the same formula:
n2 = (P2 * V2) / (R * T)
Substituting the given values:
n2 = (55 torr * 1 L) / (0.0821 L * atm / mol * K * 313.15 K)
Calculating n2:
n2 = 0.0144 mol
Since we have the same amount of water in the second flask, the total number of moles transferred to the second flask is the sum of n1 and n2:
Total moles (n_total) = n1 + n2 = 0.0396 mol + 0.0144 mol = 0.054 mol
Now, let's calculate the pressure in the second flask:
Given:
Final volume (V2) = 0.5 L
Using the ideal gas law, rearranged to solve for pressure (P2):
P2 = (n_total * R * T) / V2
Substituting the given values:
P2 = (0.054 mol * 0.0821 L * atm / mol * K * 313.15 K) / 0.5 L
Calculating P2:
P2 = 4.279 atm
Converting P2 to torr:
P2 = 4.279 atm * 760 torr / 1 atm
Calculating P2 in torr:
P2 = 3254 torr
Therefore, the pressure set up in the second flask would be 3254 torr.
To find the pressure in the second flask, we can use the combined gas law. The combined gas law states that the ratio of the initial pressure to the final pressure is equal to the ratio of the initial volume to the final volume, when the temperature remains constant.
Combining this with Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases, we can calculate the pressure in the second flask.
Here are the steps to find the pressure in the second flask:
1. Convert the initial volume to liters:
The initial volume is given as 1 L.
2. Convert the final volume to liters:
The final volume is given as 0.5 L.
3. Set up the equation:
According to the combined gas law, we have:
(P1 x V1) / T1 = (P2 x V2) / T2
In this case, the temperature remains constant, so we can simplify the equation to:
(P1 x V1) = (P2 x V2)
4. Substitute the given values:
P1 = 760 Torr (total pressure)
V1 = 1 L (initial volume)
V2 = 0.5 L (final volume)
Plugging in these values, we get:
(760 Torr x 1 L) = (P2 x 0.5 L)
5. Solve for P2:
Divide both sides of the equation by 0.5 L:
P2 = (760 Torr x 1 L) / 0.5 L
P2 = 760 Torr x 2
P2 = 1520 Torr
6. Consider the partial pressure of water vapor:
Since there is a drop or two of water in the flask, we need to account for its contribution to the total pressure in the initial flask. The aqueous tension at 40°C is given as 55 Torr.
Therefore, the pressure in the second flask, taking into account the partial pressure of water vapor, can be calculated as:
P2 (with water vapor) = P2 (without water vapor) + Pressure of water vapor
P2 (with water vapor) = 1520 Torr + 55 Torr
P2 (with water vapor) = 1575 Torr
So, the pressure set up in the second flask, including the contribution from water vapor, would be approximately 1575 Torr.