Question

What volume did a helium-filled balloon have at 16 °C and 3.03 atm if its new volume was 85.3 ml at 3.67 atm and 4.7 °C?

a. 75.2 ml

b. 90.4 ml

c. 69.8 ml

d. 82.1 ml

Answer 1

The original volume of the helium-filled balloon, using the combined gas law, was calculated to be 82.1 ml when the initial temperature and pressure were 16 °C and 3.03 atm, respectively.

Step-by-step explanation:

To solve for the original volume of the helium-filled balloon, we can make use of the combined gas law which relates pressure, volume, and temperature of a gas. The combined gas law is PV/T = k, where P is the pressure, V is the volume, T is the temperature in Kelvin, and k is a constant.

Given the final conditions of the helium-filled balloon: a volume of 85.3 ml at a pressure of 3.67 atm and a temperature of 4.7 °C (which is 277.85 K after converting from Celsius to Kelvin), we can express the final state of the gas as V2, P2, and T2. The initial conditions given are 16 °C (which is 289.15 K) and 3.03 atm, with the initial volume being unknown V1. Using the combined gas law, we have:
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for V1, we have:
V1 = (P2 * V2 * T1) / (P1 * T2)
Substituting the given values:
V1 = (3.67 atm * 85.3 ml * 289.15 K) / (3.03 atm * 277.85 K)
V1 ≈ 82.1 ml

Therefore, the original volume of the helium-filled balloon was 82.1 ml.