A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?
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To find the pressure at 273 K, we can use the ideal gas law equation:
PV = nRT
where:
P is the pressure
V is the volume (assumed to be constant for a rigid container)
n is the number of moles of gas
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
First, we need to determine the number of moles of gas. However, this information is not provided in the given question. Therefore, we cannot directly calculate the pressure at 273 K without additional information.
If we assume that the number of moles of gas remains constant, we can use the combined gas law equation to calculate the pressure at a different temperature. The combined gas law equation is derived from the ideal gas law and allows us to compare the initial and final conditions of a gas sample:
(P1 * V1) / T1 = (P2 * V2) / T2
In this case, we can rearrange the equation to solve for P2 (the pressure at 273 K):
P2 = (P1 * V1 * T2) / (V2 * T1)
Given:
P1 = 340 kPa (pressure at 713 K)
T1 = 713 K
T2 = 273 K
Since the container is rigid, the volume (V) remains constant, so we can cancel out the V terms:
P2 = (P1 * T2) / T1
Now, we can substitute the given values into the equation:
P2 = (340 kPa * 273 K) / 713 K
Simplifying the equation gives:
P2 = 130.7 kPa
Therefore, the pressure at 273 K is approximately 130.7 kPa.