Question

The minimum pressure required to compress 20 L of a gas at 1 bar to 15 L at 298 K is

Answer 1

Answer:

According to ideal gas equation, P1V1=P2V2

P1= 1 bar, V1= 20L, V2= 15L

Then P2= 1x20/15 =1.333 bar

Answer 2

Step by step explanation:

Given: Pressure 1 $= 1 bar$, Volume 1 $= 20L$ and Volume 2 $= 15L$

To find: Pressure 2.

For calculation of Pressure 2,

We know that according to the ideal gas equation,

⇒ $P1$×$V1$$=P2$×$V2$

⇒ $1$×$20 = P2$ × $15$

⇒ $P2=\frac{20}{15}$

⇒ $P2=\frac{4}{3}$ $= 1.33$

The minimum pressure required to compress is $1.33 bar$.