The gas in a fixed container was at STP before is was lifted by a balloon into the atmosphere. What would the new pressure be in kPa if the temperature at a higher altitude was 268 K?
a
.00138 kPa
b
103 kPa
c
99.4 kPa
d
.982 kPa
(P1/T1) = (P2/T2)
To find the new pressure in kPa, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure,
V is the volume (which remains constant in this case as it is a fixed container),
n is the number of moles (which also remains constant),
R is the ideal gas constant,
and T is the temperature.
Since the problem states that the gas was at STP (Standard Temperature and Pressure), we can assume that the initial pressure is 1 atmosphere or 101.3 kPa, and the initial temperature is 273 K.
Now we can rearrange the ideal gas law equation to solve for the new pressure:
P = (nRT) / V
Since n, R, and V are constant, we can simplify the equation to:
P = constant * T
Therefore, we can see that the pressure of the gas is directly proportional to the temperature. As the temperature increases, the pressure will also increase.
In this case, the initial temperature is 273 K, and the new temperature is 268 K. Since the new temperature is lower than the initial temperature, the pressure will decrease.
Given the answer choices, we can eliminate options b, c, and d since they're greater than the initial pressure of 101.3 kPa. This leaves us with option a: 0.00138 kPa as the correct answer, as it is the only choice that is significantly lower than the initial pressure.
So, the correct answer is option a: 0.00138 kPa.