What would be the new volume if the Pressure on 60L is increased from 90KPa to 150KPa? Name the law used (Ans =36L)

Question

1. What would be the new volume if the Pressure on 60L is increased from 90KPa to 150KPa? Name the law used

(Ans =36L)

Answer 1

Given:

V₁=60 L

P₁=90 kPa

V₂=?

P₂=150 kPa

Combined gas law:

(P₁V₁)÷(n₁T₁)=(P₂V₂)÷(n₂T₂)

Where

P=pressure

V=volume

n=moles

T=temperature (in Kelvin, T_K=T_C+273.15)

Now, the moles (n) and temperature (T) are constant, so eliminate those from the equation. This results in:

(P₁V₁)=(P₂V₂)

This is technically called Boyle's law, which doesn't matter unless your prof says it does.

Substitute the values in:

(90kPa)(60L)=(150kPa)(x L)

x L = (90kPa)(60L)÷(150 kPa)

x L = 5400 kPaL ÷ 150 kPa

x L = 36 L

Check: is your answer correct?

(90)(60)=(150)(36)

5400=5400

yes, the answer is correct

The law used:

Constant moles, constant temperature, variable pressure, variable volume

P₁V₁=P₂V₂

Boyle's Law