A piston is holding 125 mL of gases at a temperature of 300 K and a pressure of 2.39 atm. What is the volume of the gases if the temperature is increased to 500K and the pressure falls to 2.15 atm?

Its 231

(P1V1/T1) = (P2V2/T2)

To calculate the volume of the gases when the temperature and pressure change, we need to use the ideal gas law equation:

PV = nRT

Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature in Kelvin

To solve for the unknown volume, we can rearrange the equation to V = (nRT) / P.

Given:
P1 = 2.39 atm (initial pressure)
P2 = 2.15 atm (final pressure)
T1 = 300 K (initial temperature)
T2 = 500 K (final temperature)
V1 = 125 mL = 0.125 L (initial volume)

First, we need to convert the initial volume to liters, so V1 = 0.125 L.

We can start by calculating the number of moles of gas at the initial state using the ideal gas law equation:

n = (P1 * V1) / (R * T1)

Substituting the values, we get:

n = (2.39 atm * 0.125 L) / (0.0821 L.atm/mol.K * 300 K)

Simplifying, we find:

n ≈ 0.0122 mol

Now we can calculate the final volume using the ideal gas law equation:

V2 = (n * R * T2) / P2

Substituting the values, we get:

V2 = (0.0122 mol * 0.0821 L.atm/mol.K * 500 K) / 2.15 atm

Simplifying, we find:

V2 ≈ 0.239 L

Therefore, the volume of the gases when the temperature is increased to 500 K and the pressure falls to 2.15 atm is approximately 0.239 liters.