A gas sample occupies 6.2 L at a pressure of 201 kPa. What volume will it occupy if the pressure is increased to 335 kPa? The temperature remains the same.

Boyle's law states that the volume occupied by a fixed mass of gas is inversely proportional to the pressure, provided temperature is kept constant. That is,

PV=K

P1V1 =P2V2

P1= 201 kpa

V1 = 6.2L

P2= 335kpa

V2= ?

Making V2 the subject of formula,

V2= P1V1/P2

V2= 201× 6.2/335

V2= 1246.2/336

V2= 3.72L

Thus, the volume of the gas decreases as the pressure increases. Hope this helps, thanks

whitneytr12 whitneytr12 · Answer 2 · 2020-12-10T01:52:54+01:00

Answer:

The volume will be 3.72 L.

Explanation:

We can find the new volume by using the Ideal Gas Law:

[tex] PV = nRT [/tex]

Where:

P: is the pressure

V: is the volume

n: is the number of moles

R: is the gas constant

T: is the temperature

Initially, we have:

[tex]P_{1}V_{1} = nRT[/tex] (1)

with V₁ = 6.2 L, P₁ = 201 kPa and nRT = constant

When the pressure is increased we have:

[tex] P_{2}V_{2} = nRT [/tex] (2)

with V₂ =?, P₂ = 335 kPa and nRT = constant

By equating (1) and (2) we have:

[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]

[tex] V_{2} = \frac{P_{1}V_{1}}{P_{2}} = \frac{201 kPa*6.2 L}{335 kPa} = 3.72 L [/tex]

Therefore, the pressure is increased to 335 kPa the volume will be 3.72 L.

I hope it helps you!