Question

for many purposes we can treat butane $(c_{4}h_{10})$ as an ideal gas at temperatures above its boiling point of $-1.^{circ }c$.
suppose the pressure on a $3.0m^{3}$ sample of butane gas at $41.0^{circ }c$ is reduced to one - third its initial value.
is it possible to change the temperature of the butane at the same time such that the volume of the gas doesnt change?
if you answered yes, calculate the new temperature of the gas. round your answer to the nearest $^{circ }c$.

Explanation:

Step1: Convert initial temperature to Kelvin

The ideal gas law is \(PV = nRT\). Given \(V\) and \(n\) (amount of gas) are constant (since volume doesn't change and no gas is added/removed). So, \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\).
Initial temperature \(T_1=(41.0 + 273.15)\text{K}=314.15\text{K}\). Let initial pressure be \(P_1\) and final pressure \(P_2=\frac{1}{3}P_1\).

Step2: Use the gas - law relationship

From \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), substituting \(P_2 = \frac{1}{3}P_1\) gives \(\frac{P_1}{314.15}=\frac{\frac{1}{3}P_1}{T_2}\).
Cross - multiply: \(P_1\times T_2=\frac{1}{3}P_1\times314.15\). Since \(P_1
eq0\) (gas has non - zero pressure), we can cancel \(P_1\) out. So, \(T_2=\frac{314.15}{3}\text{K}\approx104.72\text{K}\).

Step3: Convert final temperature to Celsius

\(T_2=(104.72 - 273.15)^{\circ}\text{C}\approx - 168^{\circ}\text{C}\)

Answer:

yes, \(-168^{\circ}\text{C}\)