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?
A. 1.650atm
B. 0.825atm
C. 0.413atm
D. 0.275atm
QUICK ANSWER…
C
DETAILS…
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)
(0.825/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.
KEY-POINTS…
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 /