To find the initial volume of the gas, we can use the ideal gas law, which is expressed as:

$PV = nRT$

However, since we have the same number of moles of gas before and after the expansion, we can apply the combined gas law, which is given by:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Where:

• $P_1$ = initial pressure
• $V_1$ = initial volume
• $T_1$ = initial temperature (in Kelvin)
• $P_2$ = final pressure
• $V_2$ = final volume
• $T_2$ = final temperature (in Kelvin)

Given Data:

• $P_1 = 0.850 \, \text{atm}$
• $T_1 = 61.0 \, \text{C} = 61.0 + 273.15 = 334.15 \, \text{K}$
• $P_2 = 0.600 \, \text{atm}$
• $V_2 = 94.0 \, \text{mL}$
• $T_2 = 44.0 \, \text{C} = 44.0 + 273.15 = 317.15 \, \text{K}$

Rearranging the Equation

We need to find $V_1$:

$V_1 = V_2 \cdot \frac{P_1}{P_2} \cdot \frac{T_1}{T_2}$

Substituting Values

$V_1 = 94.0 \, \text{mL} \cdot \frac{0.850 \, \text{atm}}{0.600 \, \text{atm}} \cdot \frac{334.15 \, \text{K}}{317.15 \, \text{K}}$

Calculating:

1. Calculate $\frac{0.850}{0.600} \approx 1.4167$
2. Calculate $\frac{334.15}{317.15} \approx 1.052$
3. Multiply:

$V_1 = 94.0 \cdot 1.4167 \cdot 1.052 \approx 139.51 \, \text{mL}$

Significant Figures

The initial volume should be reported with three significant figures (as the least precise measurement is $94.0 \, \text{mL}$, which has three significant figures):

$V_1 \approx 140 \, \text{mL}$

Final Answer

The initial volume of the gas was approximately 140 mL.

Do you want more details or have any questions? Here are some related questions you might consider:

1. How does the ideal gas law relate to real gases under different conditions?
2. What assumptions are made when using the ideal gas law?
3. How does temperature affect gas pressure and volume?
4. Can you explain the concept of absolute temperature in relation to gas laws?
5. How would changes in the number of moles affect this calculation?

Tip: Always convert temperature to Kelvin when using gas laws.