An ideal gas with an initial volume of 2.30L is cooled to 12.00 Celsius where its final volume is 1.85L. What was the temperature initially?

To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of an ideal gas:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature

Rearranging the equation to solve for T1:

T1 = (P1 * V1 * T2) / (P2 * V2)

Now let's plug in the given values:

P1 is not given, but since we are assuming the gas is ideal, we can assume it remains constant throughout the process, so we can cancel it out in the equation.

V1 = 2.30 L (Given initial volume)
T2 = 12.00 Celsius (Given final temperature)
V2 = 1.85 L (Given final volume)

Let's convert Celsius to Kelvin:

T1 = (P1 * 2.30 * (12.00 + 273.15)) / (P2 * 1.85)

We still need the value of P1 to find T1, but we can't determine it without additional information. So, unfortunately, we cannot find the initial temperature without knowing the initial pressure.

To solve this problem, we can use the combined gas law, which states:

P₁V₁/T₁ = P₂V₂/T₂

where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature

In this problem, the pressure is constant since the gas is ideal, so we can simplify the combined gas law to:

V₁/T₁ = V₂/T₂

Now we can plug in the given values:

V₁ = 2.30 L
V₂ = 1.85 L
T₂ = 12.00 °C = 12.00 + 273.15 K = 285.15 K

Now we can rearrange the equation to solve for T₁:

(T₁ * V₂) / V₁ = T₂

T₁ = (T₂ * V₁) / V₂
T₁ = (285.15 K * 2.30 L) / 1.85 L

Calculating this value:

T₁ = 354.08 K

So the initial temperature of the gas was approximately 354.08 Kelvin (or Celsius, since Celsius and Kelvin have the same increment size).