# Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker.containing liquid nitrogen at -78.5°C. What will the volume of the sample of air become (at constant pressure)?

Mar 21, 2017

Use Charles law

#### Explanation:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Where V = volume
T = temperature

${V}_{1} = 1.3 L$

${T}_{1} = {24.7}^{o} C$

${T}_{2} = - {78.5}^{o} C$

Convert these units to the appropriate units if required

${24.7}^{o} C = {24.7}^{o} C + 273.15 K = 297.85 K$

$- {78.5}^{o} C + 273.15 K = 194.65 K$

Plug in the variables

$\frac{1.3 L}{297.85 K} = \frac{{V}_{1}}{194.65 K}$

$0.00436 = \frac{{V}_{1}}{194.65 K}$

${V}_{1} = 0.00436 \cdot 194.65 K$

= 0.848674L (The balloon will shrink very much)