Question

For many purposes we can treat propane (CH) as an ideal gas at temperatures above its boiling point of -42. C. Suppose the temperature of a sample of propane gas is lowered from 14.0 °C to -16.0 °C, and at the same time the pressure is changed. If the initial pressure was 1.3 atm and the volume decreased by 45.0%, what is the final pressure? Round your answer to 2 significant digits. 1 atm

Answer 1

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According to ideal gas equation we have,

PV = nRT

Where, P is pressure of the gas, V is the volume of the gas, n is no.of moles, R is gas constant and T is temperature.

Therefore, no.of moles of the gas is given by,

PI n = RT

No.of moles remains constant in this case, hence we can say that:

PVí RT P2V2 RT

PV EV, Ti Т)

Given that,

Initial temperature,

Ti = 14°C = 14 + 273 K = 287 K

Final temperature,

T, = -16°C = -16 + 273 K = 257 K

Initial pressure,

Pi=1.3 atm

Let V be the initial volume, Hence final volume will be 0.55 V

Substituting in the equation Final pressure is given by:

PiViT :: P2 = 1.3 atm XV 287 K 257 K 0.55V 1

1.3 atm 257 K :: P2 = 287 K * 0.55

ANSWER: P = 2.12 atm

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