Question

use the following information to answer the next two questions. a sample of octane (c8h18 (g)) gas in a non - rigid container occupies a volume of 6.00 l at a certain temperature and pressure. what will be its volume, in litres, in each of the following situations? show work and calculations. (2 marks) written response 2. both the pressure and the kelvin temperature are doubled. your answer: written response 3. if both the pressure and kelvin temperature are cut in half. your answer:

Explanation:

Step1: Recall the ideal gas law

The ideal gas law is $PV = nRT$. Let the initial state be $P_1V_1=nRT_1$ and the final state be $P_2V_2=nRT_2$. We can express the ratio $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$.

Step2: Solve for question 2

Given $P_2 = 2P_1$, $T_2=2T_1$ and $V_1 = 6.00$ L. Substitute into $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$:
\[

$$\begin{align*} \frac{P_1\times6.00}{T_1}&=\frac{2P_1\times V_2}{2T_1}\\ \frac{P_1\times6.00}{T_1}&=\frac{P_1\times V_2}{T_1}\\ V_2&=6.00\text{ L} \end{align*}$$

\]

Step3: Solve for question 3

Given $P_2=\frac{1}{2}P_1$, $T_2 = \frac{1}{2}T_1$ and $V_1 = 6.00$ L. Substitute into $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$:
\[

$$\begin{align*} \frac{P_1\times6.00}{T_1}&=\frac{\frac{1}{2}P_1\times V_2}{\frac{1}{2}T_1}\\ \frac{P_1\times6.00}{T_1}&=\frac{P_1\times V_2}{T_1}\\ V_2&=6.00\text{ L} \end{align*}$$

\]

Answer:

1. 6.00 L
2. 6.00 L