## Essential University Physics: Volume 1 (3rd Edition)

a) $561.73 \ J$ b) $-466.4 \ J$ c) $93.35 \ J$ d) 17 percent e) The maximum temperature is 487 Kelvin. The minimum temperature is 405 Kelvin.
a) We first find the initial temperature by simplifying the ideal gas law: $T = \frac{PV}{nR}=\frac{(8\times101.3\times10^3))(1\times10^{-3})}{.2\times8.314}=487 \ K$ Thus, we find the heat absorbed is: $Q=-W=-(-nRTln(\frac{V_2}{V_1}))$ $Q=nRTln(\frac{V_2}{V_1})$ $Q=(.2)(8.314)(487)ln2=561.73 \ J$ b) We use a similar process, this time considering heat absorbed, to find: $Q=-nRTln(\frac{V_2}{V_1})$ $Q=-nR(\frac{PV}{nR})ln(\frac{V_2}{V_1})$ $Q=-PVln(\frac{V_2}{V_1})$ $Q=-(2.05\times101.3\times10^3)(3.24\times10^{-3}ln(.5))=-466.4 \ J$ c) We add the answers to a and b to find: $W=-466.4+561.73=93.35 \ J$ d) $e=\frac{93.35}{561.73 }\times 100=17$% e) We found in part a that the maximum temperature is 487 Kelvin. The minimum temperature is: $T = \frac{PV}{nR}=\frac{(2.05\times101.3\times10^3)(3.24\times10^{-3})}{.2\times8.314}=404.64 \ K$