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Sagot :
Since the pressure varies directly with the temperature and inversely with the volume we can set the following equation:
[tex]P=\frac{kT}{V}\text{.}[/tex]We know that when T=18 K and V=27 m³, P=78 bar, therefore solving the above equation for k we get:
[tex]\begin{gathered} 78\text{ bar =k}\frac{18K}{27m^3}, \\ k=\frac{78\text{bar}27m^3}{18K}=117\frac{barm^3}{K}\text{.} \end{gathered}[/tex]Therefore, when T=9 K, V=3 m³ the pressure is:
[tex]P=117(\frac{9}{3})\text{ bar=351 bar.}[/tex]Answer: 351 bar.
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