Question

The air in a 4.50 L tank has a pressure of

2.00 atm . What is the final pressure, in atmospheres, when the air is placed in tanks that have the following volumes, if there is no change in temperature and amount of gas?

Part A- 1.12 L Express your answer using three significant figures.

PartB-2600. mLExpress your answer using four significant figures.

Part C-590. mLExpress your answer using three significant figures.

Part D-9.10 LExpress your answer using three significant figures.

Answer 1

A)

Given:

Pi = 2.00 atm

Vi = 4.50 L

Vf = 1.12 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 4.5 L = Pf * 1.12 L

Pf = 8.0357 atm

Answer: 8.04 atm

B)

Given:

Pi = 2.00 atm

Vi = 4.5 L

Vf = 2600.0 mL

= (2600.0/1000) L

= 2.6 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 4.5 L = Pf * 2.6 L

Pf = 3.4615 atm

Answer: 3.462 atm

C)

Given:

Pi = 2.00 atm

Vi = 4.5 L

Vf = 590.0 mL

= (590.0/1000) L

= 0.59 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 4.5 L = Pf * 0.59 L

Pf = 15.3 atm

Answer: 15.3 atm

D)

Given:

Pi = 2.00 atm

Vi = 4.50 L

Vf = 9.10 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 4.5 L = Pf * 9.1 L

Pf = 0.989 atm

Answer: 0.989 atm