The pressure changes from 760mmHg to 742 mmHG. When that occurs, the volume changes from 2.8L to 3.1L. The initial tempature is 20 celcious. What is the new tempature if miles remain the same? When I did this I subtracted 760 to 742 to get 18 and subtracted 2.8 to 3.1 to get .3. Then, I turned 20 to 293 because I converte it to kelvin after that I multipled .08206 to 293 to get 24.04. Then, I multiplied .3 to 18 to get 5.4. Finally, I divided over 24.04 to get .221.
You did a lot of work, most of which is not right. Here is what you do.
Use (P1V1/T1) = (P2V2/T2)
P1 is 760 mm Hg
P2 is 742 mm Hg
V1 = 2.8 L
V2 = 3.1 L
T1 = 273+20 = 293 K
T2 = ?. Solve for this and the answer will be in units of kelvin.
By the way, note the correct spelling of celsius. You made a typo in miles. I'm sure you meant mols.
To find the new temperature, you need to use the combined gas law equation, which is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Let's break down the steps to solve this problem correctly:
1. Convert the initial temperature from Celsius to Kelvin: The formula to convert Celsius to Kelvin is T(K) = T(°C) + 273.15. In this case, T(°C) = 20, so T(K) = 20 + 273.15 = 293.15 K.
2. Substitute the values into the combined gas law equation: P1 = 760 mmHg, V1 = 2.8 L, P2 = 742 mmHg, V2 = 3.1 L, and T1 = 293.15 K.
(760 mmHg * 2.8 L) / (293.15 K) = (742 mmHg * 3.1 L) / (T2)
3. Solve for T2: Cross-multiply and solve for T2.
(760 * 2.8 * T2) = (742 * 3.1 * 293.15)
(2128 * T2) = (2299.82 * 293.15)
T2 = (2299.82 * 293.15) / (2128)
T2 ≈ 318.546 K
4. Convert the new temperature from Kelvin back to Celsius: Simply subtract 273.15 from the value obtained in step 3.
T(°C) ≈ 318.546 - 273.15
T(°C) ≈ 45.396 °C
Therefore, the new temperature, with the volume remaining the same, would be approximately 45.396 °C.