Question

The air in a 6.00 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 1.20 L Express your answer using three significant figures. P1 P 1 =      atm   SubmitRequest Answer Part B 2300. mL Express your answer using four significant figures. P2 P 2 =   atm   SubmitRequest Answer Part C 550. mL Express your answer using three significant figures. P3 P 3 =   atm   SubmitRequest Answer Part D 6.60 L Express your answer using three significant figures. P4 P 4 =   atm   SubmitRequest Answer Provide Feedback

Answer 1

A)

Given:

Pi = 2.00 atm

Vi = 6.00 L

Vf = 1.20 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 6.0 L = Pf * 1.2 L

Pf = 10.0 atm

Answer: 10.0 atm

B)

Given:

Pi = 2.00 atm

Vi = 6.0 L

Vf = 2300.0 mL

= (2300.0/1000) L

= 2.3 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 6.0 L = Pf * 2.3 L

Pf = 5.2174 atm

Answer: 5.217 atm

C)

Given:

Pi = 2.00 atm

Vi = 6.0 L

Vf = 550.0 mL

= (550.0/1000) L

= 0.55 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 6.0 L = Pf * 0.55 L

Pf = 21.8182 atm

Answer: 21.8 atm

D)

Given:

Pi = 2.00 atm

Vi = 6.00 L

Vf = 6.60 L

use:

Pi*Vi = Pf*Vf

2.00 atm * 6.0 L = Pf * 6.6 L

Pf = 1.8182 atm

Answer: 1.82 atm