Part 1- The work done by the gas during this process will be 5.65 ×10⁻³ kJ.
Part 2-The heat added to the gas during this process will be 5.65 ×10×10⁻³ kJ.
<h3>What is work done by the gas?</h3>
Work is the product of pressure p and volumes V during a volume change for such a gas. The work seems to be the area under the curve that indicates how the state changes.
The work done under the isothermal process is;
![\rm W= P_1V_1 log_e(\frac{P_1}{P_2} )\\\\ W= 7 \times 1.71429 \times 10^{-3} log_e(\frac{7 }{2} )\\\\\ W= 0.0044 \ kJ](https://tex.z-dn.net/?f=%5Crm%20W%3D%20P_1V_1%20log_e%28%5Cfrac%7BP_1%7D%7BP_2%7D%20%29%5C%5C%5C%5C%20W%3D%207%20%5Ctimes%201.71429%20%20%5Ctimes%2010%5E%7B-3%7D%20log_e%28%5Cfrac%7B7%20%7D%7B2%7D%20%29%5C%5C%5C%5C%5C%20W%3D%200.0044%20%5C%20%20kJ)
For the isothermal process;
ΔU=0
![\rm \triangle Q = \triangle E + \triangle W \\\\ Q = 0 + 5.65 \times 10^{-3}\\\\ Q = 5.65 \TIMES 10^{-3} \ kJ](https://tex.z-dn.net/?f=%5Crm%20%5Ctriangle%20Q%20%3D%20%5Ctriangle%20E%20%2B%20%5Ctriangle%20W%20%5C%5C%5C%5C%20Q%20%3D%20%200%20%2B%205.65%20%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5C%20Q%20%3D%205.65%20%5CTIMES%2010%5E%7B-3%7D%20%5C%20kJ)
Hence, the work done, and the heat added by the gas during this process will be 5.65 ×10⁻³ kJ and 5.65 ×10×10⁻³ kJ respectively.
To learn more about work done by the gas, refer to the link;
brainly.com/question/12539457
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