Question

You have a 20.43L sample of helium in a weather balloon. At ground level, the conditions are: T = 21.39°C, P = 1.067atm. You release the balloon and it rises to an altitude where the temperature has dropped to 4.01°C and the pressure is 0.8185atm. You also found that there was a small leak in your balloon and you have lost 7.125% of the helium gas that was originally in the balloon. What is the volume of the balloon? Report your answer in liters to 2 decimal places.

Answer 1

answer: According to the given information , Here first we have to used the ideal gas equations

PV = NRT

1.067 * 20.43 = n * 8.314 * 294.54

n = 0.0089 mol of He

and we know that moles = mass / molar mass

mass of helium = molar mass * moles = 4* 0.0089 = 0.0356 g of He

now 7.125 % of 0.0356 = 0.0025365 g of He is lost

so , the final mass of he = 0.033 g of He is finally in ballon

then moles = 0.033 / 4 = 0.00825 mol of He

Now using PV = nRT we get

0.8185 V = 0.00825 * 8.314 * 277.16

V = 23.22 L

Hence the required answer is 23.22 L . Thank you :)