12) The gas compresses, and the piston lowers
13) The work done by the gas is [tex]7.8\cdot 10^5 J[/tex]
Explanation:
12)
For a gas kept at constant pressure, the work done on the gas is given by:
[tex]W=-p\Delta V = -p(V_f - V_i)[/tex]
where
p is the pressure of the gas
[tex]V_f[/tex] is the final volume of the gas
[tex]V_i[/tex] is the initial volume of the gas
We see that when work is done ON the gas, the work done is positive: therefore, this means that the term
[tex]\Delta V = V_f - V_i[/tex]
must be negative. This implies that
[tex]V_f < V_i[/tex]
So, the gas has being compressed (by the piston, which lowers down), and its final volume is smaller than the initial volume.
13)
Here we have a gas kept at a constant pressure of
[tex]p=1.3\cdot 10^5 Pa[/tex]
And its initial volume is
[tex]V_i = 6.0 m^3[/tex]
The gas is expanded to twice its initial volume, so the final volume is
[tex]V_f = 2 V_i = 12.0 m^3[/tex]
The work done BY the gas is
[tex]W=p\Delta V = p(V_f - V_i)[/tex]
Where here we removed the negative sign because this is the work done BY the gas. Substituting, we find
[tex]W=(1.3\cdot 10^5)(12.0-6.0)=7.8\cdot 10^5 J[/tex]
Learn more about gases here:
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