A 38.91 L sample of a qas at 165.7 K and exactly one atm compressed to 12.7 L an

Question

A 38.91 L sample of a qas at 165.7 K and exactly one atm compressed to 12.7 L and Its temperature changed to 304.3 K. What is the final pressure (in atm)? Report your answer to two (2) decimal places. Do not include units (atm) in Your Answer Answer A unknown with a heat of vaporization of 32.4 kJ/mol has a vapor pressure of 0.360 atm at 253.7 K. What is the normal boiling point(in K) of the unknown ? report your answer to one (1) decimal place. Do not include units(K) in your answer. Your Answer Answer

Explanation / Answer

V1 = 38.91 L

T1 = 165.7 K

P1 = 1 atm

V2 = 12.70

T2 = 304.3

P2 = ?

Apply ideal gas laws:

P1V1/T1 = P¨2V2/T2

P2 = P1*(V1/V2)*(T2/T1)

P2 = 1*38.91/12.7*304.3/165.7 = 5.626 atm

2)

H = 32.4 kJ/mol

P = 0.34atm T = 253.7

find normal T

Apply clasius clapeyron

ln(P2/P1) = H/R(1/T1-1/T2)

ln(1/0.34) = 32400/8.314*(1/253.7 -1/T2)

solve for T2

t2 = 272K = -0.13°C

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