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?
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Correct Answer: Option A
Explanation:
Using the combined gas law formula,
Given that P\(_1\) = 0.825, V\(_1\) = 2.76 L, V\(_2\) = 1.38 L, P\(_2\) = ?nd T1 = T2; we would have P\(_2\) = [ P\(_1\) X V\(_1\) ] ÷ V\(_2\)
: [0.825 X 2.76] ÷ 1.38
= 1.650atm
Using the combined gas law formula,
Given that P\(_1\) = 0.825, V\(_1\) = 2.76 L, V\(_2\) = 1.38 L, P\(_2\) = ?nd T1 = T2; we would have P\(_2\) = [ P\(_1\) X V\(_1\) ] ÷ V\(_2\)
: [0.825 X 2.76] ÷ 1.38
= 1.650atm