Question

A marshmallow has an initial volume of 0.068 L at standard pressure (1.0

atm). If the marshmallow is placed in a vacuum chamber and the final
volume is 0.296L, what is the pressure inside the chamber?

Answer 1

Answer: 0.232 atm

Step-by-step explanation:

The pressure inside the chamber can be calculated using the Ideal Gas Law, which states that PV = nRT.

P = pressure (atm)

V = volume (L)

n = number of moles (assumed to be constant)

R = gas constant (0.0821 L·atm/mol·K)

T = temperature (K)

Since we know the initial and final volume of the marshmallow, we can use the equation:

(P1)(V1) = (P2)(V2)

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

We can plug in the given information:

(1.0 atm)(0.068 L) = (P2)(0.296 L)

Solving for P2, we get:

P2 = (1.0 atm)(0.068 L) / (0.296 L) = 0.232 atm

So the pressure inside the chamber is 0.232 atm.