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:

1. Option A is incorrect.
2. Option B is incorrect.
3. C is correct.
4. 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 /