Question

A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the gas cools at constant pressure to 14 ∘C? How much work does the gas do in this process?

Answer 1

a) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:

`p_i V_i = nRT_i`
where

`p_i=1.0 atm=1.01 \cdot 10^5 Pa` is the initial pressure of the gas

`V_i` is the initial volume of the gas

`n=2.3 mol` is the number of moles

`R=8.31 J/K mol` is the gas constant

`T_i=240^(\circ)C=513 K` is the initial temperature of the gas

By re-arranging this equation, we can find
`V_i`:

`V_i = (nRT_i)/(p_i) = ((2.3 mol)(8.31 J/mol K)(513 K))/(1.01 \cdot 10^5 Pa)=0.097 m^3`

2) Now the gas cools down to a temperature of

`T_f = 14^(\circ)C=287 K`
while the pressure is kept constant:
`p_f = p_i = 1.01 \cdot 10^5 Pa`, so we can use again the ideal gas law to find the new volume of the gas

`V_f = (nRT_f)/(p_f)= ((2.3 mol)(8.31 J/molK)(287 K))/(1.01 \cdot 10^5 Pa) = 0.054 m^3`

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:

`W=p \Delta V= p(V_f -V_i)`
by using the data we found at point 1) and 2), we find

`W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J`
where the negative sign means the work is done by the surrounding on the gas.