Explanation:
The given data is as follows.
P = 3 atm
= [tex]3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}[/tex]
= [tex]3.03975 \times 10^{5} Pa[/tex]
[tex]V_{1}[/tex] = 9 L = [tex]9 \times 10^{-3} m^{3}[/tex] (as 1 L = 0.001 [tex]m^{3}[/tex]),
[tex]V_{2}[/tex] = 15 L = [tex]15 \times 10^{-3} m^{3}[/tex]
Heat energy = 800 J
As relation between work, pressure and change in volume is as follows.
W = [tex]P \times \Delta V[/tex]
or, W = [tex]P \times (V_{2} - V_{1})[/tex]
Therefore, putting the given values into the above formula as follows.
W = [tex]P \times (V_{2} - V_{1})[/tex]
= [tex]3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})[/tex]
= 1823.85 Nm
or, = 1823.85 J
As internal energy of the gas [tex]\Delta E[/tex] is as follows.
[tex]\Delta E[/tex] = Q - W
= 800 J - 1823.85 J
= -1023.85 J
Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.
