Please help 1. If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be?

2. I have added 15 L of air to a balloon at sea level (1.0 atm). If I take the balloon with me to Denver, where the air pressure is 0.85 atm, what will the new volume of the balloon be?

3. I’ve got a car with an internal volume of 12,000 L. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.0 atm to 1.4 atm?

number 1 is P^2=1.75

number 2 17.65 L

number 3 V2 = 8571.43 L

Step-by-step explanation:

Number 3: Given: V1 = 12000 L

P1 = 1.0 atm           P2 = 1.4 atm

Required: V2

Solution:

Use Boyle's law

(P1) × (V1) = (P2) × (V2)

V2 = [(P1) × (V1)] / (P2)

V2 = (1.0 atm × 12000L) / (1.4 atm)

V2 = 8571.43 L

Number 2: - Use Boyle's Law

- Rearrange formula (V2 = V1P1/P2)

- Put in the values and solve

Number 1: P(1)*V(1)=P(2)*V(2)

1.5 atm*5.6 L= P(2)*4.8 L

P(2)= 1.5 atm*5.6 L/4.8 L =1.75 atm

sorry, i don't know i just need points, i'm new here. chao

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.