MCAT Question about gas laws

Question

A researcher inflates a flexible balloon with 2 moles of an ideal gas at a temperature of 300 K. They then add 2 more moles of the gas and heat the balloon to 600 K. Assuming the external pressure remains constant, by what factor does the balloon's volume change?

Answer 1

We can use the ideal gas law here to answer this question. The problem explicitly states the gas is ideal, meaning the particles have no volume, there are no intermolecular interactions, and the collisions are elastic. Additionally, the external pressure "remains constant" and the balloon is flexible indicating that it can expand or retract freely.

Now that we have established that we can use the ideal gas law, we remember that it is PV=nRT where:

P is equal to pressure in Pascals

V = volume in liters (cubic meters)

n = moles of substance

R = gas constant of 8.314 J/mol*K

T = temperature in Kelvin

In this scenario, since the pressure is constant, we can simplify the equation to V=nRT. Additionally, since R is a constant, we can say that the volume is proportional to both the number of moles and the temperature of V = RT.

We can say V1 is the volume of the gas at the beginning and V2 is the volume of gas after the number of moles and temperature are doubled. This can be expressed as:

V1 = nT and V2 = 2n*2T

V2 can be further simplified as V2 = 4nT.

If we try and see how V1 and V2 are related, we see that V2 = 4*V1. This means the volume of the balloon after these changes described will change by a factor of 4.

Answer 2

This is a direct application of the Ideal Gas Law: PV=nRT. Because it's a "flexible balloon" at "constant pressure," we can see that Volume (V) is directly proportional to the number of moles (n) and the temperature (T). You doubled the moles (from 2 to 4) and you doubled the temperature (from 300K to 600K). The total change is the product of these factors: 2×2=4.