Question

A sample of O2 gas occupies a volume of 871 mL at 25 °C. If pressure remains constant, what would be the new volume if the temperature changed to:
New volume
(a) -5 °C
(b) 95 OF
(c) 1095 K

Answer 1

Answer:

$\boxed{\text{(a) 783 L; (b) 900. mL; (c) 3.20 L}}$

Explanation:

The pressure and the number of moles are constant, so, to calculate the volume, we can use Charles' Law.

$\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}$

(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

$\begin{array}{rcl}\dfrac{V_{1}}{T_{1}} &= &\dfrac{V_{2}}{T_{2}}\\\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{268.15}\\\\2.921 &= &\dfrac{V_{2}}{268.15}\\\\{ V_{2}} &=& 2.9210 \times 268.15\\&=& \textbf{783 mL}\\\end{array}\\\text{The gas will occupy \large \boxed{\textbf{783 mL}} }$

(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

$\begin{array}{rcl}\dfrac{871}{298.15} &= &\dfrac{V_{2}}{308.15}\\\\2.921 &= &\dfrac{V_{2}}{308.15}\\\\{ V_{2}} &=& 2.9210 \times 308.15\\&=& \textbf{900. mL}\\\end{array}\\\text{The gas will occupy \large \boxed{\textbf{900. mL}} }$

(c) Volume at 1095 K

$\begin{array}{rcl}\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{1095}\\\\2.921 &= &\dfrac{V_{2}}{1095}\\\\{ V_{2}} &=& 2.9210 \times 1095\\&=& \text{3200 mL}\\&=& \textbf{3.20 L}\\\end{array}\\\text{The gas will occupy \large \boxed{\textbf{3.20 L}} }$