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The initial gas pressure is 6 xx 10^6Pa and the volume 1 m^3. Expansion at constant temperature leads to its volume being increased two-fold. Using numerical methods calculate the work of expansion of the gas. Compare with the formula in xi 27.6 and estimate the error. |
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Answer» Solution :Divide the change in volume into 10 equal PARTS so that `DeltaV = 0.1 m^3`. The element of work along a short path is `W_i- bar(p_i) DeltaV`. (Fig.) .Therefore the total work is `W = DeltaV (bar(p_1) + bar(p_2) +… + bar(p_10))`  write the data in the form of a table (Table) . It follows that hte work done by a gas expanding at constant temperature is `W = 0.1 xx 41.58 xx 10^5 = 4.16 xx 10^5 J` USING the FORMULA of `xi 27.6`, we have `= 2.3 xx 6 xx 0.301 xx 10^5 = 4.17 xx 10^5 J` The RELATIVE error in the numerical calculation is `epsilon = (0.001 xx 100%)/(4.17) = 0.24%`. 
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