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Chapter 10: Problem 109

An ideal gas at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of \(0.800 \mathrm{~L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is \(92.66 \mathrm{kPa}\), what is the volume of the bulb that was originally filled with gas?

Short Answer

Expert verified
The initial volume of the bulb filled with gas is approximately 0.583 L.

Step by step solution

01

Write down the initial and final states of pressure and volume

Before the stopcock is opened, the pressure is given as 152 kPa and the volume is unknown. We will denote the volume of the initial gas-filled bulb as V1. After the stopcock is opened, the final pressure is 92.66 kPa, and the total volume is the sum of the volume of the initial bulb (V1) and the evacuated bulb (0.800 L). We can denote the final volume as V2. Initial state: Pressure, P1 = 152 kPa, Volume, V1 (unknown) Final state: Pressure, P2 = 92.66 kPa, Volume, V2 = V1 + 0.800 L
02

Use the ideal gas law to create an equation

Since the temperature is constant, we can write the ideal gas law for both the initial and final states, noting that the amount of gas, n, is also constant: Initial state: \(P_1V_1 = nRT\) Final state: \(P_2V_2 = nRT\)
03

Solve for V1 using the conservation of volume principle

Divide the equation of the final state by the equation of the initial state: \(\frac{P_2V_2}{P_1V_1} = \frac{nRT}{nRT}\) The right side of the equation simplifies to 1, since n, R, and T are all constants: \(\frac{P_2V_2}{P_1V_1} = 1\) Now we can plug in the given values for P2, P1, and V2: \(\frac{92.66 (V_1 + 0.800)}{152 V_1} = 1\)
04

Solve for V1

We can now solve for V1 by multiplying both sides by \(152V_1\): \(92.66 (V_1 + 0.800) = 152 V_1\) Divide both sides by 92.66: \(V_1 + 0.800 = \frac{152}{92.66} V_1\) Solve for V1: \(V_1(1 - \frac{152}{92.66}) = 0.800\) \(V_1 = \frac{0.800}{1 - \frac{152}{92.66}}\) Calculating the expression, we get: \(V_1 \approx 0.583 L\) So the initial volume of the bulb filled with gas is approximately 0.583 L.

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Most popular questions from this chapter

Complete the following table for an ideal gas: $$\begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & ? \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & ? \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & ? \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ ? \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array}$$

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