If the quantity of oxygen occupying 2.76L container at a…
If the quantity of oxygen occupying 2.76L container at a pressure of 0.825 atm and 300k is reduced by one-half, what is the pressure exerted by the remaining gas?
Please observe that the volume of the container and the temperature remained constant, only the quantity (moles) and the pressure are changing. We have a formula in mind:
PV = n RT
P = n RT/V
Therefore, P/n = RT/V
P/n = constant
Where P = pressure,
T = temperature,
V = volume,
R = gas constant, and
n = number of moles also taken as the quantity.
Now, since P/n = constant, it implies that P1/n1 = P2/n2
Where P1 is given as 0.825atm
P2 = unknown = ?
Also, let n1 = A
And since the quantity was said to have reduced by half, n2 = A/2 =
Substituting these values into the equation, we have:
P1/n1 = P2/n2
0.825/A = P2/(A/2)
A) × ( A/2) = P2
Therefore, P2 = 0.825/2 = 0.4125 = approximately 0.413atm
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is incorrect.
- C is correct.
- D is not the correct answer.
You may please note these/this:
- There is a short cut, since P and n are directly proportional, a decrease in P will also cause a decrease in n in the same measure.
- A decrease in n by half will trigger a decrease in P by the same half, hence, we can quickly see that P2 = half of P1.
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/ culled from 2020 JAMB-UTME chemistry past question 13 /
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