After 273 m3 of ethylene oxide is formed at 748 kPa and 525 K, the gas is cooled at constant volume to 293 K.The new pressure is _____kPa
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To find the new pressure of the ethylene oxide gas after it is cooled at constant volume, we can make use of the combined gas law:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 = initial pressure of the gas (given as 748 kPa)
V1 = initial volume of the gas (given as 273 m3)
T1 = initial temperature of the gas (given as 525 K)
P2 = new pressure of the gas (to be determined)
V2 = final volume of the gas (which remains constant)
T2 = final temperature of the gas (given as 293 K)
Since the volume remains constant, V1 = V2, we can simplify the equation to:
P1 / T1 = P2 / T2
Now we can substitute the given values into the equation:
(748 kPa) / (525 K) = P2 / (293 K)
To find P2, we can cross-multiply and solve for P2:
(748 kPa) * (293 K) = P2 * (525 K)
P2 = (748 kPa * 293 K) / (525 K)
Calculating the above expression:
P2 = 418,444 kPa
Therefore, the new pressure of the ethylene oxide gas after cooling at constant volume to 293 K is approximately 418,444 kPa.
To find the new pressure after cooling the gas at constant volume, we can make use of the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas (constant in this case)
n is the number of moles of gas
R is the gas constant
T is the temperature of the gas
We can rearrange the equation to solve for the final pressure (P2) in terms of the initial pressure (P1) and the initial and final temperatures (T1 and T2):
P2 = (P1 * T2) / T1
Given:
Initial volume (V) = 273 m^3
Initial pressure (P1) = 748 kPa
Initial temperature (T1) = 525 K
Final temperature (T2) = 293 K
Now, let's calculate the new pressure (P2):
P2 = (P1 * T2) / T1
= (748 kPa * 293 K) / 525 K
Calculating this expression:
P2 = (219364 kPa * K) / 525 K
= 418 kPa (approximately)
Therefore, the new pressure after cooling the gas at constant volume is approximately 418 kPa.