Answer:
[tex]\boxed{\text{(a) 783 L; (b) 900. mL; (c) 3.20 L}}[/tex]
Explanation:
The pressure and the number of moles are constant, so, to calculate the volume, we can use Charles' Law.
[tex]\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}[/tex]
(a) Volume at -5 °C
Data:
V₁ = 871mL; T₁ = 25 °C
V₂ = ?; T₂ = -5 °C
Calculations:
(i) Convert temperatures to kelvins
T₁ = ( 25 + 273.15) K = 298.15 K
T₂ = (-5 + 273.15) K = 268.15 K
(ii) Calculate the new volume
[tex]\begin{array}{rcl}\dfrac{V_{1}}{T_{1}} &= &\dfrac{V_{2}}{T_{2}}\\\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{268.15}\\\\2.921 &= &\dfrac{V_{2}}{268.15}\\\\{ V_{2}} &=& 2.9210 \times 268.15\\&=& \textbf{783 mL}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{783 mL}}$}[/tex]
(b) Volume at 95 °F
95 °F = (95 - 32) × 5/9 = 63 × 5/9 = 35 °C
35 °C = (35 + 273.15) K = 308.15 K
[tex]\begin{array}{rcl}\dfrac{871}{298.15} &= &\dfrac{V_{2}}{308.15}\\\\2.921 &= &\dfrac{V_{2}}{308.15}\\\\{ V_{2}} &=& 2.9210 \times 308.15\\&=& \textbf{900. mL}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{900. mL}}$}[/tex]
(c) Volume at 1095 K
[tex]\begin{array}{rcl}\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{1095}\\\\2.921 &= &\dfrac{V_{2}}{1095}\\\\{ V_{2}} &=& 2.9210 \times 1095\\&=& \text{3200 mL}\\&=& \textbf{3.20 L}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{3.20 L}}$}[/tex]
