A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa a nd the new volume is 55.0 L. What was the initial pressure of the sample? Show your work, using the G.U.E.S.S. method.
1 answer:
Answer:
71.92 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
T1 = 50°C = 50 + 273 = 323K
V1 = 105L
T2 = -25°C = -25 + 273 = 248K
P2 = 105.4 kPa
P1 = ?
V2 = 55.0 L
Using P1V1/T1 = P2V2/T2
P1 × 105/323 = 105.4 × 55/248
105P1/323 = 5797/248
0.325P1 = 23.375
P1 = 23.375 ÷ 0.325
P1 = 71.92 kPa
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<h3>Further explanation</h3>
Given
12.6 g of Oxygen
Required
mass of KCl
Solution
Reaction
2KClO3 ⇒ 2KCl + 3O2
mol O2 :
= mass : MW
= 12.6 : 32 g/mol
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