Question

A balloon of air now occupies 10.0 L at 25.0 C and 1.00 atm. What was the initial temperature, in celsius , if it occupied 9.40 L and was in the freezer with a pressure of 0.939?

Answer 1

Using the combined gas law, the initial temperature of the air in the balloon when it was in the freezer is calculated to be 6.54 C.

Step-by-step explanation:

To find the initial temperature of the balloon in the freezer, we can use the combined gas law, which includes Boyle's Law, Charles's Law, and Gay-Lussac's Law. The combined gas law is stated as: (P1 * V1) / T1 = (P2 * V2) / T2 where P is pressure, V is volume, and T is temperature in Kelvins.

We initially have the following conditions: V1 = 9.40 L, V2 = 10.0 L, P2 = 1.00 atm, T2 = 25.0 C = 298.15 K (since Kelvin is Celsius + 273.15), and P1 = 0.939 atm. We want to find T1, the initial temperature in Celsius.

Rearranging the combined gas law to solve for T1 gives us: T1 = (P1 * V1 * T2) / (P2 * V2).

Now, substituting the values, we have: T1 = (0.939 atm * 9.40 L * 298.15 K) / (1.00 atm * 10.0 L) = 279.69 K.

Converting this to Celsius: T1 in Celsius = 279.69 K - 273.15 = 6.54 C.

Therefore, the initial temperature of the balloon when it was in the freezer was 6.54 C.