An ideal gas is taken through a quasi-static process described by P = aV2, with a...

 

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An ideal gas is taken through a quasi-static process described by P = aV2, with a = 2.25 atm/m6, as shown in the figure. The gas is expanded to twice its original volume of 1.00 m. How much work is done on the expanding gas in this process? P P= av 1.00 m 2.00 m Part 1 of 3 - Conceptualize The final pressure is four times the initial pressure. Looking at the graph, we can estimate the average pressure during expansion to be half the final pressure. An atmosphere of pressure is equal to 1.013 105 Pa, so we estimate the work to have a magnitude on the order of 106 J. Part 2 of 3 - Categorize The work done on the gas is the negative of the area under the curve P = av from V, to V. The work done on the gas is negative, meaning that the expanding gas does positive work. We will find the amount of work by integrating work between the initial and final volumes. Part 3 of 3 - Analyze The work done on the gas is the negative of the area under the curve P = av between V, and V. W = - -I" PdV VE av dv= Substituting V, 2V = 2(1.00 m) = 2.00 m, we have = Jatm/m)(D W = - x 106 J. Submit Skip (you cannot come back) 105 Pa/atm)[( ]m)] An ideal gas is taken through a quasi-static process described by P = aV2, with a = 2.25 atm/m6, as shown in the figure. The gas is expanded to twice its original volume of 1.00 m. How much work is done on the expanding gas in this process? P P= av 1.00 m 2.00 m Part 1 of 3 - Conceptualize The final pressure is four times the initial pressure. Looking at the graph, we can estimate the average pressure during expansion to be half the final pressure. An atmosphere of pressure is equal to 1.013 105 Pa, so we estimate the work to have a magnitude on the order of 106 J. Part 2 of 3 - Categorize The work done on the gas is the negative of the area under the curve P = av from V, to V. The work done on the gas is negative, meaning that the expanding gas does positive work. We will find the amount of work by integrating work between the initial and final volumes. Part 3 of 3 - Analyze The work done on the gas is the negative of the area under the curve P = av between V, and V. W = - -I" PdV VE av dv= Substituting V, 2V = 2(1.00 m) = 2.00 m, we have = Jatm/m)(D W = - x 106 J. Submit Skip (you cannot come back) 105 Pa/atm)[( ]m)]