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Chapter 1: Problem 2

State whether the following statements are true or false. When a \(200 \mathrm{ml}\) gas, in a closed vessel, is heated to \(50^{\circ} \mathrm{C}\) from \(25^{\circ} \mathrm{C}\), the volume becomes doubled.

Short Answer

Expert verified
False

Step by step solution

01

1. Write down the initial parameters

First, write down the given initial parameters: Initial volume (\(V_1\)) = \(200 \,\mathrm{ml}\) Initial temperature (\(T_1\)) = \(25^{\circ}\mathrm{C}\) Final temperature (\(T_2\)) = \(50^{\circ}\mathrm{C}\)
02

2. Convert the temperature to Kelvin scale

To work with the ideal gas law, we need to convert the given temperatures to Kelvin scale. \(T_1\) in Kelvin = \(25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\) \(T_2\) in Kelvin = \(50^{\circ}\mathrm{C} + 273.15 = 323.15\,\mathrm{K}\)
03

3. Use the ideal gas law relationship

We assume the gas behaves ideally and the pressure remains constant. In this situation, we can use Charles' Law, which states that \((V_1/T_1) = (V_2/T_2)\) where, \(V_2\) is the final volume. We need to find \(V_2\) and see if it is twice \(V_1\).
04

4. Solve the equation to find the final volume

Now, solve for \(V_2\) using the given parameters: \((V_1/T_1) = (V_2/T_2)\) \(\frac{200\,\mathrm{ml}}{298.15\,\mathrm{K}} = \frac{V_2}{323.15\,\mathrm{K}}\) Multiply both sides by \(323.15\,\mathrm{K}\) to isolate \(V_2\): \(V_2 = \frac{(200\,\mathrm{ml})(323.15\,\mathrm{K})}{298.15\,\mathrm{K}}\) Now calculate \(V_2\): \(V_2 \approx 216.43\,\mathrm{ml}\)
05

5. Compare the final volume with the doubled initial volume

We found that the final volume of the gas, \(V_2\), is approximately \(216.43\,\mathrm{ml}\). Now, let's check if this value is equal to double the initial volume: Doubled initial volume = \(2 \times 200\,\mathrm{ml} = 400\,\mathrm{ml}\) Since \(216.43\,\mathrm{ml} \ne 400\,\mathrm{ml}\), the statement is false.

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