Explanation:
The given data is as follows.
n = 2, [tex]V_{1} = 0.02 m^{2}[/tex], [tex]V_{2} = 0.05 m^{2}[/tex]
[tex]T_{1} = 0.02 m^{2}[/tex] = 300 K , [tex]T_{2}[/tex] = 500 K
P = 1 bar
Equation for work done will be as follows.
W = [tex]-P \times \Delta V[/tex]
= [tex]-1 bar \times (0.05 m^{2} - 0.02 m^{2})[/tex]
= - 3000 J
Hence, formula for heat added is as follows.
Q = [tex]nC_{p} \Delta T[/tex]
Putting given values into the above formula as follows.
Q = [tex]nC_{p} \Delta T[/tex]
= [tex]2 \times \frac{7}{2} \times (500 K - 300 K)[/tex]
= 11639.6 J
Thus, we can conclude that the amount of heat added is 11639.6 J.
