Question

1. A sample of gas has a volume of 6.20 L at 20°C at a pressure of 1.10 atm. What is its volume at the same temperature and at a pressure of 0.925 atm? (Hint: What equation will you use if the temperature is constant?).

Answer 1

Answer:

7.37 L is the new volume

This can be understood as a system that maintains its constant T ° and if the moles of the gas are not modified, the pressure varies inversely proportional to the volume.

Explanation:

We can apply the Ideal Gases Law, to solve this problem. Moles are the same in both cases, the T° also keeps on constant.

P . V = n . R . T      For this first situation

P . V = n . R . T      For the second situation

V = n . R . T / P

n . R . T will be the same, because n is the number of moles, T means T° and R is a physic constant, so for the first case we can define:

P₁ . V₁, and for the second case, P₂ . V₂

We match both equations, and we replace data

6.20L. 1.10 atm = 0.925 atm . V₂

(6.20L. 1.10 atm) / 0.925 atm = V₂

V₂ = 7.31 L

If the volume decreases, the pressure increases more because the molecules collide more with the walls of the container but the opposite happens where if the volume increases (as in the case of exercise), the pressure will be lower. Gas molecules will not collide so much with the walls

Answer 2

Answer:

The new volume will be 7.37 L

Explanation:

Step 1: Data given

Volume = 6.20 L

Pressure = 1.10 atm

Temperature will stay constant

Pressure will change to 0.925 atm

Step 2:

Since The temperature, the number of moles and the gas constant will be constant, we will use the following equation

p1*V1 = p2*V2

⇒with p1 = the initial pressure = 1.10 atm

⇒with V1 = the initial volume = 6.20 L

⇒with p2 = the new pressure = 0.925 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.10 * 6.20 = 0.925 * V2

V2 = 7.37 L

The new volume will be 7.37 L