A Sample of propane, C3H8, occupies 5.00L at 270C and 309torr. The volume of the canister is changed to 1.85L while the pressure remains constant. What is the Temperature, in K and 0C, of the gas after the change?

What would be the setup for this problem?
Thank you.

(V1/T1) = (V2/T2)

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To solve this problem, you can use the combined gas law equation, which relates the initial and final conditions of a gas when pressure, volume, and temperature change while keeping the amount of gas constant.

The combined gas law equation is given as follows:

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

Where:
P1 and P2 are the initial and final pressures respectively,
V1 and V2 are the initial and final volumes respectively,
T1 and T2 are the initial and final temperatures respectively.

In this case, the problem states that the pressure remains constant, so the initial pressure (P1) is equal to the final pressure (P2). Let's label the initial volume (V1) as 5.00 L, the final volume (V2) as 1.85 L, and the initial temperature (T1) as 27°C. We need to find the final temperature (T2) in both Kelvin (K) and Celsius (°C).

The setup for this problem would be as follows:

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

Substituting the values into the equation:

(P1 * 5.00 L) / (27°C + 273.15) = (P1 * 1.85 L) / T2

Since the pressures cancel out, the equation simplifies to:

(5.00 L) / (27°C + 273.15) = (1.85 L) / T2

Now, we can solve for T2 by rearranging the equation as follows:

T2 = (1.85 L * (27°C + 273.15)) / 5.00 L

Finally, calculate T2 using the above equation to find the temperature in Kelvin. To convert it to °C, subtract 273.15 from the Kelvin temperature.

Note: It's important to always convert temperatures to Kelvin when using the gas laws because temperature must be in Kelvin in these calculations.