A closed inexpansible vessel contains air saturated with water vapour at 77 degree Celsius.The total pressure in the vessel is 1007mmHg.Calculate the new pressure if the temperature is reduced at 27 degree Celsius (S.V.P of water at 77degree celcius and 27 degree Celsius respectively are 314mmHg and 27mmHg. Treat the air in the vessel as an ideal gas
work this as two problems; air and water.
the air is at 77C and 1007-314 kpa
new air is at P-27 at 27C.
but in each case the moles of air are the same.
n=PV/RT
so P1V/RT1=P2V/RT2 or
P1/T1=P2/T2 where P2=P-27
so
(1007-314)/(77+273)=(P-27)/(27+373) so solve for P, the new total pressure.
Notice this assumes the condensed water does not change the air volume appreciably.
Well, it seems like we got a little science experiment going on here! So, we have a closed vessel with some hot and steamy air inside. But now we want to cool things down a bit. Let's see what happens when we do that!
First, let's take a look at the initial situation. We have air saturated with water vapor at 77 degrees Celsius. The total pressure in the vessel is 1007 mmHg. That's a pretty steamy situation if you ask me!
Now, when we bring the temperature down to 27 degrees Celsius, we need to consider the vapor pressures of water at these two temperatures. At 77 degrees Celsius, the saturation vapor pressure (S.V.P) of water is 314 mmHg. And at 27 degrees Celsius, it's only 27 mmHg.
Since we're treating the air in the vessel as an ideal gas, we can make use of the ideal gas law to calculate the new pressure. The ideal gas law states that the pressure times the volume divided by the temperature is equal to a constant.
Let's call the initial pressure P1, the initial temperature T1, and the final pressure P2, and the final temperature T2.
Using the ideal gas law, we can set up the following equation:
(P1 * V) / T1 = (P2 * V) / T2
Since we're keeping the volume constant, we can cancel it out.
Now, let's plug in the values we know. We have P1 = 1007 mmHg, T1 = 77 degrees Celsius, P2 = ?, and T2 = 27 degrees Celsius.
So, we have:
(1007 mmHg * V) / 77 degrees Celsius = (P2 * V) / 27 degrees Celsius
Now, let's cross multiply and solve for P2:
(1007 mmHg * V * 27 degrees Celsius) = (77 degrees Celsius * P2 * V)
We can cancel out the volume again:
1007 mmHg * 27 degrees Celsius = 77 degrees Celsius * P2
Now, let's solve for P2:
P2 = (1007 mmHg * 27 degrees Celsius) / 77 degrees Celsius
P2 = 352 mmHg
So, the new pressure in the vessel, after cooling down to 27 degrees Celsius, would be 352 mmHg.
That's a significant drop in pressure! Just like my enthusiasm when I'm stuck in a steamy situation and someone turns on the air conditioning!
To solve this problem, we can use the Ideal Gas Law, which states
PV = nRT
Where:
P = Pressure
V = Volume (which remains constant in this case as the vessel is closed and inexpansible)
n = number of moles of gas
R = Ideal Gas Constant
T = Temperature in Kelvin
Since the volume remains constant, we can say:
P1/T1 = P2/T2
Given:
P1 = 1007 mmHg
T1 = 77 °C = 350 K
T2 = 27 °C = 300 K
S.V.P of water at 77 °C = 314 mmHg
S.V.P of water at 27 °C = 27 mmHg
To find P2, we first need to find the pressure due to only water vapor at 77 °C, and then subtract it from the total pressure.
Using the equation P1/T1 = P2/T2, we can find the pressure due to water vapor at 77 °C:
P2 = (P1/T1) * T2
P2 = (1007/350) * 300
P2 ≈ 861 mmHg
Now we can find the pressure due to the air alone:
P_air = P1 - P2
P_air = 1007 - 861
P_air ≈ 146 mmHg
Finally, we can add the saturated vapor pressure at 27 °C to find the new total pressure:
P2_new = P_air + S.V.P at 27 °C
P2_new = 146 + 27
P2_new ≈ 173 mmHg
Therefore, the new pressure in the vessel after reducing the temperature to 27 °C is approximately 173 mmHg.
To calculate the new pressure, we can use the combined gas law, which states that the ratio of the initial pressure to the new pressure is equal to the ratio of the initial temperature to the new temperature, assuming the volume and amount of gas remain constant.
The combined gas law equation is as follows:
(P1/T1) = (P2/T2)
Where:
P1 = initial pressure
T1 = initial temperature
P2 = new pressure
T2 = new temperature
Given:
P1 = 1007 mmHg
T1 = 77 degrees Celsius
T2 = 27 degrees Celsius
S.V.P of water at 77 degrees Celsius = 314 mmHg
S.V.P of water at 27 degrees Celsius = 27 mmHg
First, we need to calculate the partial pressure of the water vapor at both temperatures. The partial pressure is the pressure exerted by a component in a gas mixture.
Partial pressure due to water vapor at 77 degrees Celsius:
P1_water = S.V.P of water at 77 degrees Celsius = 314 mmHg
Partial pressure due to water vapor at 27 degrees Celsius:
P2_water = S.V.P of water at 27 degrees Celsius = 27 mmHg
To find the pressure exerted by air alone at 77 degrees Celsius, we subtract the partial pressure of water vapor from the total pressure:
P_air1 = P1 - P1_water
= 1007 mmHg - 314 mmHg
= 693 mmHg
Using the combined gas law, we can now calculate the new pressure (P2) when the temperature is reduced to 27 degrees Celsius:
(P1_air / T1) = (P2_air / T2)
Substituting the known values:
(693 mmHg / 77°C) = (P2_air / 27°C)
Now, solving for P2_air:
P2_air = (693 mmHg / 77°C) * 27°C
≈ 242.25 mmHg
Finally, to find the new pressure (P2), we add the partial pressure of water vapor at 27 degrees Celsius to the pressure exerted by air alone:
P2 = P2_air + P2_water
≈ 242.25 mmHg + 27 mmHg
≈ 269.25 mmHg
Therefore, the new pressure when the temperature is reduced to 27 degrees Celsius is approximately 269.25 mmHg.