6.52 × 10⁴ L. (3 sig. fig.)
Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."
Consider the ideal gas law:
[tex]P\cdot V = n\cdot R\cdot T[/tex],
where
The question is asking for the final volume [tex]V[/tex] of the gas. Rearrange the ideal gas equation for volume:
[tex]V = \dfrac{n \cdot R \cdot T}{P}[/tex].
Both the temperature of the gas, [tex]T[/tex], and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.
Start with the absolute temperature of the gas:
The volume of the gas is proportional to its temperature if both [tex]n[/tex] and [tex]P[/tex] stay constant.
[tex]V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}[/tex].
Now, keep the temperature at [tex]T_1 =258.65\;\text{K}[/tex] and change the pressure on the gas:
The volume of the gas is proportional to the reciprocal of its absolute temperature [tex]\dfrac{1}{T}[/tex] if both [tex]n[/tex] and [tex]T[/tex] stays constant. In other words,
[tex]V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}[/tex]
(3 sig. fig. as in the question.).
See if you get the same result if you hold [tex]T[/tex] constant, change [tex]P[/tex], and then move on to change [tex]T[/tex].
