Question

The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C?

Answer 1

Answer: 0.529 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.998 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 2 L

$V_2$ = final volume of gas = 3.5 L

$T_1$ = initial temperature of gas = $34.0^oC=273+34.0=307.0K$

$T_2$ = final temperature of gas = $12.0^oC=273+12.0=285.0K$

Now put all the given values in the above equation, we get:

$\frac{0.998\times 2}{307.0K}=\frac{P_2\times 3.5}{285.0K}$

$P_2=0.529atm$

Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm