Question

Helium gas with a volume of 3.50 l , under a pressure of 0.160 atm and at a temperature of 39.0 ∘c , is warmed until both pressure and volume are doubled. What is the final temperature?

Answer 1

To find the final temperature, we can use the ideal gas law equation PV = nRT. After converting the initial temperature to Kelvin, we can solve for the final temperature using the equation P1V1 / T1 = P2V2 / T2.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law equation, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the initial temperature from Celsius to Kelvin by adding 273.15. So, 39.0 ℃ + 273.15 = 312.15 K.

Next, we can solve for the final temperature, T2, using the equation. For a given amount of gas at constant pressure, we have P1V1 / T1 = P2V2 / T2. Since the pressure and volume are doubled, we can write the equation as (0.160 atm)(3.50 L) / (312.15 K) = (2 * 0.160 atm)(2 * 3.50 L) / T2.

Solving for T2, we get T2 = (0.160 atm)(3.50 L)(312.15 K) / (2 * 0.160 atm)(2 * 3.50 L).