Respuesta :

Answer:

[tex]186.01\ \text{mL}[/tex]

Explanation:

[tex]P_1[/tex] = Initial pressure = 1.47 atm

[tex]P_2[/tex] = Final pressure = [tex]10^5\ \text{Pa}[/tex]

[tex]T_1[/tex] = Initial temperature = [tex]44^{\circ}\text{C}[/tex]

[tex]T_2[/tex] = Final temperature = [tex]273.15\ \text{K}[/tex]

[tex]V_1[/tex] = Initial volume = 145 mL

[tex]V_2[/tex] = Final volume

We have the relation

[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow V_2=\dfrac{1.47\times 101325\times 145\times 273.15}{(44+273.15)\times 10^5}\\\Rightarrow V_2=186.01\ \text{mL}[/tex]

Final volume of the sample is [tex]186.01\ \text{mL}[/tex].

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