Significant figures -- wrong answer to Thermodynamics question?

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The discussion centers on the appropriate use of significant figures in a thermodynamics problem where the calculated heat, Q, is 54,500 J. There is debate about whether the final answer should be rounded to one significant figure, resulting in 60,000 J, due to the initial volume's precision. Participants highlight that the precision of chart values could influence the significant figures, with some values possibly having more significant figures than initially considered. Additionally, there is a mention of the conversion factor between atmospheres and pascals affecting the overall precision. The conversation emphasizes the importance of accurately interpreting significant figures based on the data provided.
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Homework Statement
I am confused why the answer to the problem below has more significant figures than it should.
Relevant Equations
##\Delta E_{int} = Q - W##
For this problem,
1680317487199.png

The solution is, ## Q = 54500 J ## , however, since the calculation for heat uses the first law which involves ##5 \times 10^{-3) m^3 ## value for the initial volume, should the finial answer not be to 1 sig fig as well? This would give ## 60000 J ##.

Many thanks!
 

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Whether it is ##5\cdot 10^{-3}##, ##5.0\cdot 10^{-3}##, ##5.00\cdot 10^{-3}## etc. depends on how precisely you think you can read it off the chart. Similarly the other chart readings. The more limiting precision would be the 30kJ.
 
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haruspex said:
Whether it is ##5\cdot 10^{-3}##, ##5.0\cdot 10^{-3}##, ##5.00\cdot 10^{-3}## etc. depends on how precisely you think you can read it off the chart. Similarly the other chart readings. The more limiting precision would be the 30kJ.
Thank you for your reply @haruspex!

True you make a very valid point! I had not considered that the chart values could have more sig figs that they point.
 
I think they expect you to observe that at 5 l, the pressure is 50 atm and at 12 l, the pressure is 20 atm, (exactly). They apparently took the conversion between atm and Pa to be 1.00000 rather than 101325.
 
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Chestermiller said:
I think they expect you to observe that at 5 l, the pressure is 50 atm and at 12 l, the pressure is 20 atm, (exactly). They apparently took the conversion between atm and Pa to be 1.00000 rather than 101325.
Thank you for reply @Chestermiller !
 
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