Question 8c472

Feb 25, 2014

The volume of the balloon will be 7.8 × 10⁷ L.

This is a combined gas law problem:

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

${P}_{1}$ = 1 atm = 760 mm Hg
${V}_{1}$ = 2.8 × 10⁷ L
${T}_{1}$ = 0.00 °C = 273.15 K

${P}_{2}$ = 243 mm Hg
${V}_{2}$ = ?
${T}_{2}$ = -29.0 °C = 244.2 K

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

V_2 =V_1 × P_1/P_2 × T_2/T_1 = 2.8 × 10^-7 L × (760 mm Hg)/(243 mm Hg) × (244.2 K)/(273.15 K)# = 7.8 × 10⁷ L

Note: The answer has only two significant figures because V₁ has only two significant figures.