Step-by-step explanation:
The given data is as follows.
n = 2,
,
![V_(2) = 0.05 m^(2)](https://img.qammunity.org/2020/formulas/chemistry/college/a4mnip4kxu9pa53sv58430wa5ia23vca1q.png)
= 300 K ,
= 500 K
P = 1 bar
Equation for work done will be as follows.
W =
=
= - 3000 J
Hence, formula for heat added is as follows.
Q =
![nC_(p) \Delta T](https://img.qammunity.org/2020/formulas/chemistry/college/3ny5xwkishhtzil33yflrbs02byn8qq8hn.png)
Putting given values into the above formula as follows.
Q =
![nC_(p) \Delta T](https://img.qammunity.org/2020/formulas/chemistry/college/3ny5xwkishhtzil33yflrbs02byn8qq8hn.png)
=
![2 * (7)/(2) * (500 K - 300 K)](https://img.qammunity.org/2020/formulas/chemistry/college/mlaql8ua2fb8lbe50gvdwda6feob85tne4.png)
= 11639.6 J
Thus, we can conclude that the amount of heat added is 11639.6 J.