For many purposes we can treat propane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Question

4 years ago

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For many purposes we can treat propane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

sample of propane gas is raised from ⁢ to , and at the same time the pressure is changed. If the initial pressure was and the volume increased by , what is the final pressure? Round your answer to the correct number of significant digits.

Chemistry

2 answers:

Delicious77 [7] · Answer 1 · 2021-12-20T11:40:46+03:00

Answer:

The equation you should use is:

Meaning of the initials of the equation:

INITIAL PRESSURE (P1)

FINAL PRESSURE (P2)

INITIAL VOLUME (V1)

FINAL VOLUME (V2)

FINAL TEMPERATURE (T2)

INITIAL TEMPERATURE (T1)

then the final equation would be:

(P1XV1) / T1 = (P2XV2) / T2

Explanation:

This equation is due to the fact that the gas is considered to be an ideal gas, so when behaving as such the values of "n" which is the number of moles is the same in the initial and final stage as the constant

"R" that has a value of 0.082 (with their respective units) both at the end and at the beginning of the reaction.

By not varying these components of the equation it is unnecessary to put them, since they would cancel themselves.

In the equation we mentioned before, it is necessary that if you want to know the final pressure, that is, P2, you have to clear it, considering that the final equation for this specific exercise is:

P2 = (P1XV1XT2) / V2

ryzh [129] · Answer 2 · 2021-12-20T09:16:21+03:00

Answer: (This question is missing some values).

The combined gas law is used to determine the change in volume, pressure and temperature of gases. It states that the ratio between the pressure-volume product and temperature is a constant.

Mathematically, P1V1/T1 = P2V2/T2, where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, T1 and T2 are the initial and final temperatures in kelvin.

Explanation:

The boiling point of propane is -42°C.

Assuming the initial and final temperatures are 10°C and 25°C respectively; the volume increased by 20%; if the initial pressure = 1atm, final pressure can be found using the equation P1V1/T1 = P2V2/T2.

P1 = 1atm, P2 = ?, V1 = V, V2 = 0.2V, T1 = 10 + 273K = 283K, T2 = 25 + 273K = 298K.

Making P2 subject of formula, P2 = P1V1T2/V2T1

P2 = 1 * V *298/(0.2 V * 283)