a) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
[tex]p_i V_i = nRT_i[/tex]
where
[tex]p_i=1.0 atm=1.01 \cdot 10^5 Pa[/tex] is the initial pressure of the gas
[tex]V_i[/tex] is the initial volume of the gas
[tex]n=2.3 mol[/tex] is the number of moles
[tex]R=8.31 J/K mol[/tex] is the gas constant
[tex]T_i=240^{\circ}C=513 K[/tex] is the initial temperature of the gas
By re-arranging this equation, we can find [tex]V_i[/tex]:
[tex]V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3 [/tex]
2) Now the gas cools down to a temperature of
[tex]T_f = 14^{\circ}C=287 K[/tex]
while the pressure is kept constant: [tex]p_f = p_i = 1.01 \cdot 10^5 Pa[/tex], so we can use again the ideal gas law to find the new volume of the gas
[tex]V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3[/tex]
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:
[tex]W=p \Delta V= p(V_f -V_i)[/tex]
by using the data we found at point 1) and 2), we find
[tex]W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J[/tex]
where the negative sign means the work is done by the surrounding on the gas.
