Question

A sealed balloon is filled with 1.30 L of helium at 17°C and 1.03 atm. The balloon rises to a point in the atmosphere where the pressure is 225 torr and the temperature is −31 °C. What is the change in the volume of the balloon as it ascends from 1.03 atm to a pressure of 225 torr?

Answer 1

ΔV=2.481 L

Step-by-step explanation:

As we know that

`(PV)/(T)=(P_(2)V_(2) )/(T_(2))`

Always use temperature in Kelvin and Pressure in Atm

Given Data

P₁=1.03 atm

P₂=225 torr =(225/760)=0.296 atm

V₁=1.30

T=17⁰C=290.15 K

T₂=-31⁰C=242.15 K

Substitute the given values

So

`(1.03*1.30)/(290) =(V_(2)(0.296) )/(242.15) \\1.2224*10^(-3)V_(2)=4.6217*10^(-3)\\ V_(2)=3.781`

You have to find the difference of two volumes:

So

ΔV=3.781 - 1.30

ΔV=2.481 L