Questions: For many purposes we can treat methane (CH4) as an ideal gas at temperatures above its boiling point of -161 °C. Suppose the temperature of a sample of methane gas is raised from -35.0°C to 1.0°C, and at the same time the pressure is decreased by 15.0% Does the volume of the sample increase, decrease, or stay the same? O increase If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent. %

For many purposes we can treat methane (CH4) as an ideal gas at temperatures above its boiling point of -161 °C.
Suppose the temperature of a sample of methane gas is raised from -35.0°C to 1.0°C, and at the same time the pressure is decreased by 15.0%
Does the volume of the sample increase, decrease, or stay the same? O increase
If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent. %

Transcript text: For many purposes we can treat methane $\left(\mathrm{CH}_{4}\right)$ as an ideal gas at temperatures above its boiling point of $-161 .{ }^{\circ} \mathrm{C}$. Suppose the temperature of a sample of methane gas is raised from $-35.0^{\circ} \mathrm{C}$ to $1.0^{\circ} \mathrm{C}$, and at the same time the pressure is decreased by $15.0 \%$ \begin{tabular}{|l|l|} \hline Does the volume of the sample increase, decrease, or stay the same? & O increase \\ \hline \begin{tabular}{l} If you said the volume increases or decreases, calculate the percentage change in \\ the volume. Round your answer to the nearest percent. \end{tabular} & $\square \%$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Understanding the Problem

We need to determine if the volume of a methane gas sample increases, decreases, or stays the same when the temperature is raised and the pressure is decreased. Additionally, if the volume changes, we need to calculate the percentage change in volume.

Step 2: Combined Gas Law

The combined gas law is given by: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] where $P$ is pressure, $V$ is volume, and $T$ is temperature in Kelvin.

Step 3: Converting Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin: \[ T_1 = -35.0^\circ \text{C} + 273.15 = 238.15 \text{ K} \] \[ T_2 = 1.0^\circ \text{C} + 273.15 = 274.15 \text{ K} \]

Step 4: Expressing Pressure Change

The pressure is decreased by 15.0%, so: \[ P_2 = P_1 \times (1 - 0.15) = 0.85 P_1 \]

Step 5: Applying the Combined Gas Law

Using the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Substitute $P_2 = 0.85 P_1$: \[ \frac{P_1 V_1}{238.15} = \frac{0.85 P_1 V_2}{274.15} \] Cancel $P_1$ from both sides: \[ \frac{V_1}{238.15} = \frac{0.85 V_2}{274.15} \]

Step 6: Solving for $V_2$

Rearrange to solve for $V_2$: \[ V_2 = V_1 \times \frac{274.15}{238.15 \times 0.85} \] Calculate the ratio: \[ \frac{274.15}{238.15 \times 0.85} \approx 1.355 \] Thus: \[ V_2 \approx 1.355 V_1 \]

Step 7: Calculating Percentage Change in Volume

The percentage change in volume is: \[ \text{Percentage change} = \left( \frac{V_2 - V_1}{V_1} \right) \times 100\% = (1.355 - 1) \times 100\% = 35.5\% \]