Question

for many purposes we can treat propane $(c_{3}h_{8})$ as an ideal gas at temperatures above its boiling point of $-42.^{circ}c$.
suppose the temperature of a sample of propane gas is lowered from $90.0^{circ}c$ to $54.0^{circ}c$, and at the same time the pressure is decreased by $5.0%$.
does the volume of the sample increase, decrease, or stay the same?
if you said the volume increases or decreases, calculate the percentage change in the volume. round your answer to the nearest percent.

Explanation:

Step1: Convert temperatures to Kelvin

The ideal gas law uses absolute temperature (Kelvin). The conversion formula is \(T(K)=T(^{\circ}C)+273.15\).
For the initial temperature \(T_1 = 90.0^{\circ}C+273.15 = 363.15\ K\)
For the final temperature \(T_2=54.0^{\circ}C + 273.15=327.15\ K\)
Let the initial pressure be \(P_1 = P\), then the final pressure \(P_2=(1 - 0.05)P=0.95P\)

Step2: Use the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

We want to find \(V_2\) in terms of \(V_1\). Rearranging the combined gas law formula \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)
Substitute \(P_1 = P\), \(P_2 = 0.95P\), \(T_1 = 363.15\ K\), and \(T_2 = 327.15\ K\) into the formula:
\(V_2=\frac{P\times V_1\times327.15}{0.95P\times363.15}\)
The \(P\) cancels out. Then \(V_2=\frac{327.15}{0.95\times363.15}V_1\)
\(V_2=\frac{327.15}{345.9925}V_1\approx0.9456V_1\)

Step3: Calculate the percentage change

The percentage change in volume is given by \(\%\text{change}=\frac{V_2 - V_1}{V_1}\times100\)
Substitute \(V_2\approx0.9456V_1\) into the formula: \(\%\text{change}=(0.9456 - 1)\times100=- 5.44\%\approx - 5\%\)

Answer:

decrease; \(-5\%\)