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Chapter 6: Problem 39

A sample of an ideal gas at 15.0 atm and 10.0 \(\mathrm{L}\) is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. Calculate the work in units of kJ for the gas expansion. (Hint: Boyle's law applies.)

Short Answer

Expert verified
The work done by the gas during the expansion is \(-13.169\,\text{kJ}\). The negative sign indicates that the gas has done work on the surroundings.

Step by step solution

01

List given information and formula

We are given the initial pressure (P₁), initial volume (V₁), and the constant external pressure (P₂) during expansion: - P₁ = 15.0 atm - V₁ = 10.0 L - P₂ = 2.00 atm Since this is an isothermal process, we can use Boyle's Law to relate initial and final pressures and volumes: \(P_1 V_1 = P_2 V_2\) The work formula for an ideal gas in an isothermal process: \(W = -P_{ext} (V_2 - V_1)\)
02

Use Boyle's Law to find the final volume (V₂)

To find the final volume V₂, we will use Boyle's Law: \(P_1 V_1 = P_2 V_2\) Plug in the values and solve for V₂: \(15.0 \,\text{atm} \times 10.0\,\text{L} = 2.00\,\text{atm} \times V_2\) Now, divide both sides by 2.00 atm to get V₂: \(V_2 = \frac{15.0\,\text{atm} \times 10.0\,\text{L}}{2.00\,\text{atm}} = 75.0\,\text{L}\)
03

Calculate the work done by the gas during expansion

Now that we have V₂, we can plug it into the work formula for an isothermal process: \(W = -P_{ext} (V_2 - V_1)\) Substitute the given values and the found V₂: \(W = -2.00\,\text{atm} (75.0\,\text{L} - 10.0\,\text{L})\) Calculate the difference in volumes and multiply by the negative external pressure: \(W = -2.00\,\text{atm} \times 65.0\,\text{L}\) Multiply the values to find the work done: \(W = -130\,\text{atm} \times \text{L}\)
04

Convert the work to kilojoules (kJ)

In order to convert the work from atm·L to kJ, we use the conversion factor: 1 atm·L = 101.3 J. \(W = -130\,\text{atm} \times \text{L} \times \frac{101.3\,\text{J}}{1\,\text{atm}\cdot\text{L}}\) Multiply and divide by the conversion factor: \(W = -13169\,\text{J}\) Finally, convert J to kJ by dividing by 1000: \(W = -13.169\,\text{kJ}\) The work done by the gas during the expansion is -13.169 kJ. The negative sign indicates that the gas has done work on the surroundings.

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Most popular questions from this chapter

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