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)
1465torr
To find the pressure set up in the second flask, we can use the combined gas law equation:
(P1V1) / T1 = (P2V2) / T2
Given:
P1 = 760 Torr (initial pressure in the first flask)
V1 = 1 L (initial volume in the first flask)
T1 = 40°C (initial temperature)
V2 = 0.5 L (volume in the second flask)
T2 = 40°C (final temperature)
First, we need to calculate the pressure contributed by nitrogen only, without considering the water vapor. This can be done using Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is the sum of the partial pressures of each individual gas.
P_nitrogen = P_total - P_water_vapor
Given:
P_total = 760 Torr (total pressure in the first flask)
P_water_vapor = 55 Torr (aqueous tension at 40°C)
P_nitrogen = 760 Torr - 55 Torr
P_nitrogen = 705 Torr
Now, we can substitute the values into the combined gas law equation:
(P1V1) / T1 = (P2V2) / T2
(760 Torr * 1 L) / (40°C + 273.15) = (P2 * 0.5 L) / (40°C + 273.15)
Simplifying:
760 Torr * 1 L * (40°C + 273.15) = P2 * 0.5 L * (40°C + 273.15)
P2 = (760 Torr * 1 L * (40°C + 273.15)) / (0.5 L * (40°C + 273.15))
P2 = 1520 Torr
Therefore, the pressure set up in the second flask will be 1520 Torr when all the contents are transferred.
To find the pressure in the second flask, we can use the ideal gas law. The ideal gas law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's find the number of moles of nitrogen gas in the initial flask. We know that the flask contains 1 liter of gas at 40°C and a total pressure of 760 torr. We can use the ideal gas law to find the number of moles:
PV = nRT
n = PV / RT
where P = 760 torr, V = 1 L, R is the ideal gas constant (0.0821 L atm / K mol), and T is the temperature in Kelvin (40°C + 273 = 313 K).
Substituting the values in, we get:
n = (760 torr)(1 L) / (0.0821 L atm / K mol)(313 K)
≈ 29.6 moles
Now, let's consider the water droplets in the flask. The aqueous tension at 40°C is given as 55 torr. The water vapor exerts a partial pressure, so the total pressure in the flask is the sum of the pressure exerted by the nitrogen gas and the pressure exerted by the water vapor:
Total pressure = pressure due to nitrogen gas + pressure due to water vapor
= 760 torr + 55 torr
= 815 torr
Now, we can find the pressure in the second flask. The volume of the second flask is 0.5 L. Again, we can use the ideal gas law:
PV = nRT
P = nRT / V
where n is the number of moles of the gas (which we found to be 29.6 moles), R is the ideal gas constant (0.0821 L atm / K mol), T is the temperature in Kelvin (40°C + 273 = 313 K), and V is the volume of the second flask (0.5 L).
Substituting the values in, we get:
P = (29.6 moles)(0.0821 L atm / K mol)(313 K) / 0.5 L
≈ 1530 torr
Therefore, the pressure in the second flask will be approximately 1530 torr.
P1(dry)×V1 =P2 ×V2
P1=Pwet - aqueous tension=760 - 55=705torr
V2=0.5L.HenceP2(dry gas)= 705×1/0.5=1410 torr
P2(wet gas)=1410+55=1456torr