Question

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 and 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.

Answer 1

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