Question

1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?

Answer 1

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

• P1= 16 atm
• V1
• P2= ?
• V2= V1÷2= $\frac{V1}{2}$ because the volume is halved.

So:

16 atm*V1= P2* $\frac{V1}{2}$

Solving:

$\frac{16 atm*V1*2}{V1}$=P2

16 atm*2= P2

32 atm= P2

The pressure will be 32 atm, twice the initial pressure.

2. Now

• P1= 5.5 atm
• V1
• P2= ?
• V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

$\frac{5.5 atm*V1}{3*V1}$=P2

$\frac{5.5 atm}{3}$= P2

1.83 atm= P2

The pressure will be 1.83 atm, one third of the initial pressure.