The work done by the gas is -940 kJ
Explanation:
In this process, we are told that the product of pressure and volume remains constant:
[tex]pV=const.[/tex]
so we can write
[tex]p_1 V_1 = p_2 V_2[/tex]
where
[tex]p_1 = 1.4 bar[/tex] is the initial pressure
[tex]p_2 = 6.8 bar[/tex] is the final pressure
[tex]V_1=4.25 m^2[/tex] is the initial volume
Solving for [tex]V_2[/tex], we find the final volume:
[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]
Now by looking at the equation of state of an ideal gas:
[tex]pV=nRT[/tex] (1)
we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by
[tex]W=nRTln(\frac{V_2}{V_1})[/tex]
And by looking again at (1), we can substitute (nRT) with (pV), so we get
[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]
Converting the pressure into SI units,
[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]
So the work done is
[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]
Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).
Learn more about ideal gases:
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