The work done on the given gas in one cycle is -405.3 kJ.
The given parameters;
- initial pressure of the gas, P₁ = 2 atm
- final pressure of the gas, P₂ = 6 atm
- initial volume of the gas, V₁ = 1 m³
- final volume of the gas, V₂ = 3 m³
Convert the pressure to Pascal (N/m²);
1 atm = 101325
The work done in one cycle is the area of the triangle and it is calculated as follows
[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]
the net work done on the gas = - 405.3 kJ
Thus, the work done on the given gas in one cycle is -405.3 kJ.
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