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 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?

Answer 1

Answer: The final pressure is 0.81 kPa

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.58 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v

$V_2$ = final volume of gas = $v-\frac{40}{100}\times v=0.6v$

$T_1$ = initial temperature of gas = $25^0C=(25+273)K=298K$

$T_2$ = final temperature of gas = $-22^0C=(-22+273)K=251K$

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

$\frac{0.58\times v}{298}=\frac{P_2\times 0.6v}{251}$

$P_2=0.81kPa$

The final pressure is 0.81 kPa