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 25.0°C to - 22.0 °C, and at the same time the pressure is changed. If the initial pressure was 0.58 kPa and the volume decreased by 40.0%, what is the final pressure?

Answers

Answer 1

Answer: The final pressure is 0.81 kPa

Explanation:

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 0.58 kPa

[tex]P_2[/tex] = final pressure of gas = ?

[tex]V_1[/tex] = initial volume of gas = v

[tex]V_2[/tex] = final volume of gas = [tex]v-\frac{40}{100}\times v=0.6v[/tex]

[tex]T_1[/tex] = initial temperature of gas = [tex]25^0C=(25+273)K=298K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-22^0C=(-22+273)K=251K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.58\times v}{298}=\frac{P_2\times 0.6v}{251}[/tex]

[tex]P_2=0.81kPa[/tex]

The final pressure is 0.81 kPa


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Answers

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Answers

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Answers

Answer:

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