Gas is confined in a tank at a pressure of 1.05E+6 Pa and a temperature of 14.0°C. If half of the gas is withdrawn and the temperature is raised to 63.2°C, what is the new pressure in the tank?
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PV = nRT
P = Pressure
V = Volume
n = number of gas molecules
R = Gas constant
T = Temperature
So first we have
PV = nR (14 + 273.15)
So for the second case we have
PV = (n/2)R (63.2 + 273.15)
V is constant since it's confined in a tank
R is the same in both equations
rewrite the second equation as:
PV = (1/2)nR(63.2 + 273.15)
now you can just cross out V, n, and R since they're equal in both equations and the remaining number you get on the right is the ratio of the original pressure to the new pressure.
i.e.
From the first equation:
right side is equal to 287.15
Second equation
right side is equal to 168.175
168.175/287.15 = .5856
So the new pressure is 58.56% of the original
P = 6.15E5 Pa
P = Pressure
V = Volume
n = number of gas molecules
R = Gas constant
T = Temperature
So first we have
PV = nR (14 + 273.15)
So for the second case we have
PV = (n/2)R (63.2 + 273.15)
V is constant since it's confined in a tank
R is the same in both equations
rewrite the second equation as:
PV = (1/2)nR(63.2 + 273.15)
now you can just cross out V, n, and R since they're equal in both equations and the remaining number you get on the right is the ratio of the original pressure to the new pressure.
i.e.
From the first equation:
right side is equal to 287.15
Second equation
right side is equal to 168.175
168.175/287.15 = .5856
So the new pressure is 58.56% of the original
P = 6.15E5 Pa