Questions: For many purposes we can treat ammonia (NH₃) as an ideal gas at temperatures above its boiling point of -33 °C. Suppose the temperature of a sample of ammonia gas is raised from 61.0°C to 94.0°C, and at the same time the pressure is increased by 15.0%. Does the volume of the sample increase, decrease, or stay the same? O increase O decrease If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent. O stays the same

For many purposes we can treat ammonia (NH₃) as an ideal gas at temperatures above its boiling point of -33 °C. Suppose the temperature of a sample of ammonia gas is raised from 61.0°C to 94.0°C, and at the same time the pressure is increased by 15.0%. Does the volume of the sample increase, decrease, or stay the same? O increase O decrease If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent. O stays the same

Transcript text: For many purposes we can treat ammonia $\left(\mathrm{NH}_{3}\right)$ as an ideal gas at temperatures above its boiling point of $-33 .{ }^{\circ} \mathrm{C}$. Suppose the temperature of a sample of ammonia gas is raised from $61.0^{\circ} \mathrm{C}$ to $94.0^{\circ} \mathrm{C}$, and at the same time the pressure is increased by $15.0 \%$. \begin{tabular}{|l|l|} \hline Does the volume of the sample increase, decrease, or stay the same? & O increase \\ O decrease \\ \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} & O stays the same \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Identify the Initial and Final Conditions

We are given the initial and final temperatures and the percentage increase in pressure. Let's denote:

Initial temperature: $ T_1 = 61.0^\circ \mathrm{C} = 61.0 + 273.15 = 334.15 \, \mathrm{K} $
Final temperature: $ T_2 = 94.0^\circ \mathrm{C} = 94.0 + 273.15 = 367.15 \, \mathrm{K} $
Initial pressure: $ P_1 $
Final pressure: $ P_2 = P_1 \times 1.15 $ (since the pressure is increased by 15%)

Step 2: Use the Ideal Gas Law

The ideal gas law is given by: \[ PV = nRT \] For the same amount of gas ($ n $) and the same gas constant ($ R $), we can write the relationship between the initial and final states as: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Step 3: Solve for the Volume Ratio

Rearranging the equation to solve for the volume ratio $ \frac{V_2}{V_1} $: \[ \frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1} \] Substitute $ P_2 = 1.15 P_1 $: \[ \frac{V_2}{V_1} = \frac{P_1 \times 367.15}{1.15 P_1 \times 334.15} = \frac{367.15}{1.15 \times 334.15} \]

Step 4: Calculate the Volume Ratio

Perform the calculation: \[ \frac{V_2}{V_1} = \frac{367.15}{1.15 \times 334.15} \approx \frac{367.15}{384.2725} \approx 0.9554 \]

Step 5: Determine the Percentage Change in Volume

The percentage change in volume is given by: \[ \left( \frac{V_2}{V_1} - 1 \right) \times 100\% = (0.9554 - 1) \times 100\% = -4.46\% \] Rounding to the nearest percent: \[ -4.46\% \approx -4\% \]