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  1. Problems
  2. /A gas at a steady 1.4 × 10⁵ Pa is squeezed from 8.0 to 3.5 litres…
EasyPhysicsHigh School (11–12)Heat & Thermodynamics › Laws of ThermodynamicsNumericalBoard ExamAPA-LevelIB

A gas at a steady 1.4 × 10⁵ Pa is squeezed from 8.0 to 3.5 litres…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A gas at a steady 1.4×105 Pa1.4 \times 10^{5}\ \text{Pa}1.4×105 Pa is squeezed from 8.08.08.0 to 3.53.53.5 litres. Find the work done by the gas. Who does the work, the gas or its surroundings?

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Work it out on paper, then type your number. The answer itself stays hidden.

J
Need a reminder? The key ideaShowHide

Work done by a gas is the area under its path on a pressure-volume graph. At a steady pressure W=p ΔVW = p\,\Delta VW=pΔV; along a straight line, the average pressure times ΔV\Delta VΔV; at a steady temperature W=nRTln⁡V2V1W = nRT\ln\dfrac{V_2}{V_1}W=nRTlnV1​V2​​. Expanding gives positive work, squeezing negative. 1 L=10−3 m31\ \text{L} = 10^{-3}\ \text{m}^31 L=10−3 m3.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. ΔV=(3.5−8.0)×10−3 m3=−4.5×10−3 m3\Delta V = (3.5 - 8.0) \times 10^{-3}\ \text{m}^3 = -4.5 \times 10^{-3}\ \text{m}^3ΔV=(3.5−8.0)×10−3 m3=−4.5×10−3 m3.

Step 2. W=1.4×105 Pa×(−4.5×10−3 m3)=−630 JW = 1.4 \times 10^{5}\ \text{Pa} \times (-4.5 \times 10^{-3}\ \text{m}^3) = -630\ \text{J}W=1.4×105 Pa×(−4.5×10−3 m3)=−630 J.

Answer: −630 J-630\ \text{J}−630 J: the surroundings do 630 J630\ \text{J}630 J of work on the gas.

Watch out. A squeeze always gives negative work in our convention.

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