Question

1. a helium balloon with a pressure of 1.00 atm and a volume of 4.50 l at 20.0 °c is released into the atmosphere. what volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is - 20.0 °c?
2. a given sample of gas has a volume of 4.20 l at 60.0 °c and 1.00 atm pressure. calculate its pressure if the volume changed to 5.00 l and the temperature to 27.0 °c.

Explanation:

Step1: Convert temperatures to Kelvin

The initial temperature $T_1 = 60.0^{\circ}C+273.15 = 333.15K$ and the final temperature $T_2=27.0^{\circ}C + 273.15=300.15K$. The initial volume $V_1 = 4.20L$ and initial pressure $P_1 = 1.00atm$, the final volume $V_2 = 5.00L$.

Step2: Use the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We want to find $P_2$. Rearranging the combined - gas law formula for $P_2$ gives $P_2=\frac{P_1V_1T_2}{V_2T_1}$.

Step3: Substitute the values

$P_2=\frac{1.00atm\times4.20L\times300.15K}{5.00L\times333.15K}$
$P_2=\frac{1.00\times4.20\times300.15}{5.00\times333.15}atm$
$P_2=\frac{1260.63}{1665.75}atm\approx0.757atm$

Answer:

$0.757atm$