The work done on the gas mixture during this process is the work done is [tex]$-5.3 \mathrm{~J}$[/tex]
Work done by a gas against a constant external pressure is given by the expression,
[tex]$\mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}$[/tex]
Substitute the values to find the work done during this process.
[tex]$\begin{aligned}\mathrm{W} & =-0.98 \mathrm{~atm}(800 \mathrm{~mL}-150 \mathrm{~mL}) \\& =-0.637 \mathrm{~atm} . \mathrm{L} \\& =-0.637 \times 8.3145 \mathrm{~J} \\& =-5.3 \mathrm{~J}\end{aligned}$[/tex]
The formula for work completed according to science is W = F * d. In this instance, the force acting on the block is constant, but its impact on the block's displacement and force direction are different. Here, the force F responds to the displacement d at an angle of.
Therefore, the work done is [tex]$-5.3 \mathrm{~J}$[/tex].
Complete question: The gas mixture inside one of the cylinders of an automobile engine expands against a constant external pressure of 0.98 atm, from an initial volume of 150 mL (at the end of the compression stroke) to a final volume of 800 mL. Calculate the work done on the gas mixture during this process, and express it in joules.
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