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A gas occupies 6.50L of space at 25.00∘C and a pressure of 1.60atm. The gas was moved to a 12.0L container and the new pressure was found to be 1.20atm. What is the new temperature of the gas in degrees Celsius? Use −273.15∘C for absolute zero. Report your answer to three significant figures. Provide your answer below:

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Final answer:

The new temperature of the gas when it is moved to a 12.0L container and has a pressure of 1.20atm is 73.60°C.

Step-by-step explanation:

To find the new temperature of the gas in degrees Celsius, we can use the combined gas law. The combined gas law relates pressure, volume, and temperature of a gas as follows:

P1 * V1 / T1 = P2 * V2 / T2 where P is pressure, V is volume, and T is temperature in Kelvins.

We are given the initial conditions of the gas as:

  • Initial Pressure (P1) = 1.60 atm,
  • Initial Volume (V1) = 6.50 L,
  • Initial Temperature (T1) = 25.00°C = 298.15 K (since we add 273.15 to Celsius to convert to Kelvins).

We are also given the conditions after the gas is moved:

  • Final Volume (V2) = 12.0 L,
  • Final Pressure (P2) = 1.20 atm.

Plugging in the known values to the combined gas law equation, we will solve for T2, the final temperature in Kelvins. After finding T2, we'll convert it back to degrees Celsius to get our answer.

Let's solve the equation:

(1.60 atm * 6.50 L) / 298.15 K = (1.20 atm * 12.0 L) / T2

T2 = (1.20 atm * 12.0 L * 298.15 K) / (1.60 atm * 6.50 L)

T2 ≈ 346.75 K

To find the temperature in degrees Celsius, subtract 273.15 from the Kelvin temperature: T2 in °C = 346.75 K - 273.15 = 73.60°C

Therefore, the new temperature of the gas in the 12.0 L container at 1.20 atm is 73.60°C.

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