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Chapter 7: Problem 35

A sample of an ideal gas at \(15.0 \mathrm{atm}\) and \(10.0 \mathrm{L}\) is allowed to expand against a constant external pressure of \(2.00 \mathrm{atm}\) to a volume of \(75.0 \mathrm{L}\). Calculate the work in units of \(\mathrm{kJ}\) for the gas expansion.

Short Answer

Expert verified
The work done by the gas expansion is approximately \(-13.17 \mathrm{kJ}\).

Step by step solution

01

Identify given values

We are given the following values: Initial pressure, P1 = 15.0 atm Initial volume, V1 = 10.0 L Final volume, V2 = 75.0 L Constant external pressure, P_ext = 2.00 atm
02

Calculate the change in volume

To find the change in volume, subtract the initial volume from the final volume: ∆V = V2 - V1 ∆V = 75.0 L - 10.0 L ∆V = 65.0 L
03

Convert pressure and volume to SI units

To calculate the work in joules, we need to convert the external pressure and change in volume to SI units. 1 atm = 101325 Pa 1 L = 0.001 m³ P_ext = 2.00 atm × 101325 Pa/atm = 202650 Pa ∆V = 65.0 L × 0.001 m³/L = 0.065 m³
04

Calculate the work done

Now, use the formula for work done against constant pressure: Work = -P_ext × ∆V Work = -202650 Pa × 0.065 m³ Work = -13172.25 J
05

Convert the work to kilojoules

To convert the work to kilojoules, divide by 1000: Work = -13172.25 J ÷ 1000 Work = -13.17 kJ The work done by the gas expansion is approximately -13.17 kJ.

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