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

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

Step-by-step explanation:

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

`(V_(1))/(T_(1)) = (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}(V_(1))/(T_(1)) &= &(V_(2))/(T_(2))\\\\(871)/(298.15) &= &(V_(2))/(268.15)\\\\2.921 &= &(V_(2))/(268.15)\\\\{ V_(2)} &=& 2.9210 * 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}(871)/(298.15) &= &(V_(2))/(308.15)\\\\2.921 &= &(V_(2))/(308.15)\\\\{ V_(2)} &=& 2.9210 * 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}\\(871)/(298.15) &= &(V_(2))/(1095)\\\\2.921 &= &(V_(2))/(1095)\\\\{ V_(2)} &=& 2.9210 * 1095\\&=& \text{3200 mL}\\&=& \textbf{3.20 L}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{3.20 L}}$}`