What pressure (in atm) is required to contain 0.034 moles of oxygen gas in a 5.1 L container at a temperature of 25.00C
1 answer:
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Answer:
Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-
1 mole of react with 2 moles of
0.008174 mole of react with 2*0.008174 moles of
Moles of = 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,