Question

A marshmallow has an initial volume of 0.084 L at standard pressure (1.0 atm). If the marshmallow is placed in a vacuum chamber and the final volume is 0.785L, what is the pressure inside the chamber?

Answer 1

The new pressure on the gas P2 = 0.10atm

Step-by-step explanation:

Data;

V1 = 0.084L

P1 = 1.0atm

V2 = 0.785L

P2 = ?

This question is a practical problem where Boyle's law is applied.

According to Boyle's law, the pressure of a fixed mass of gas is inversely proportional to its volume provided that the temperature on the gas remains constant

Mathematically

P = k / v, k = PV

P1*V1 = P2*V2 = P3*V3..........Pn*Vn

P1 * V1 = P2 * V2

Solving for P2,

P2 = (P1 * V1 ) / V2

P2 = (0.084 * 1) / (0.785)

P2 = 0.10atm

The new pressure of the gas is 0.10atm