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.
What is work done by the gas?
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;
[tex]\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[/tex]
For the isothermal process;
ΔU=0
[tex]\rm \triangle Q = \triangle E + \triangle W \\\\ Q = 0 + 5.65 \times 10^{-3}\\\\ Q = 5.65 \TIMES 10^{-3} \ kJ[/tex]
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;
https://brainly.com/question/12539457
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