Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

Chapter 18 - Exercises and Problems - Page 331: 47

Answer

Please see the work below.

Work Step by Step

We know that $\frac{W_{adia}}{W_{iso}}=\frac{1-(\frac{V_1}{V_2})^{\gamma}(\frac{V_2}{V_1})}{(\gamma-1)\times \ln(\frac{V_2}{V_1})}$ This simplifies to: $\frac{W_{adia}}{W_{iso}}=\frac{1-(\frac{V_1}{V_2})^{\gamma-1}}{(\gamma-1)\times ln(\frac{V_2}{V_1})}$ We plug in the known values to obtain: $\frac{W_{adia}}{W_{iso}}=\frac{1-(2)^{\frac{5}{3}-1}}{(\frac{5}{3}-1)\times \ln(\frac{1}{2})}=1.27$ (b) As we know that in an isothermal process, the change in internal energy is zero, hence the extra work by adiabatic gas is used to raise the internal energy of the gas. Both processes have similar characteristics for heat and work.

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