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3 years ago

Using the combined gas law, calculate the following. Also, what other law could you have used to solve this problem?

Chemistry

1 answer:

blsea [12.9K]3 years ago

8 0

Answer:

4.23 L

Explanation:

The question doesn't specify pressure so

i will just assume it is 1 atm. This won't affect the answer. The same result can be achieved using the combined gas law and Charle's law, so long as pressure is kept constant in the combined gas law.

Using the combined gas law

(P1 × V1)/ T1 = (P2 × V2)/T2

where

P1 = 1 atm

V1 = 4 L

T1 =293.15 K

P2 = 1 atm

V2 =?

T2=310.15 K

We can rearrange the equation to make V2 the subject

V2= (((P1 × V1)/ T1) × T2)/P2

V2 = (((1 × 4)/293.15)×310.15)/1

V2 = 4.23 L

Using Charle's law

V1/ T1 = V2/ T2

V2 = (V1/T1) × T2

V2 = (4/293.15) ×310.15

V2 =4.23L

Hopefully this is correct

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

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Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

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Answer:C

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