Question

a balloon is filled with 2.45 l helium at 1.01 atm. the balloon is then brought to an altitude at which the atmospheric pressure is only 449 mmhg. what is the final volume (in l) of the balloon? assume temperature is constant. please report just the numerical answer to 3 significant figures and do not enter units of l.

Explanation:

Step1: Convert pressure units

We know that \(1\ atm = 760\ mmHg\). So, \(P_1=1.01\ atm\) and \(P_2 = 449\ mmHg=\frac{449}{760}\ atm\approx0.591\ atm\), \(V_1 = 2.45\ L\).

Step2: Apply Boyle's law (\(P_1V_1=P_2V_2\))

Rearrange the formula for \(V_2\): \(V_2=\frac{P_1V_1}{P_2}\).
Substitute the values: \(V_2=\frac{1.01\times2.45}{0.591}\).
Calculate \(1.01\times2.45 = 2.4745\).
Then \(V_2=\frac{2.4745}{0.591}\approx4.19\).

Answer:

4.19