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Two moles of helium gas (gamma=5/3)are initially at temperature 27^@ C and occupy a volume of 20litres.The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. What are the final volume and pressureof the gas ? |
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Answer» <P> Solution : At point B,Pressure `P'=P=2.5xx10^5 N//m^2` , `V'=2V=40 xx10^(-3)m^3` As pressure is CONSTANT in the process AB, making its volume doubled, its temperature will also be doubled. Thus temperature at point B is T'= 600 K The gas now undergoes adiabatic expansion to cool down to T''=T=300K We know for an adiabatic process `TV^(gamma-1)` =constant `T'(V')^(gamma-1)=T''(T'')^(gamma-1)` `((V'')/(V'))=((T')/(T''))^(1//gamma-1) =(600/300)^(1//(5/3-1))` `=(2)^(3//2)=2sqrt2` Thus final volume is `V''=(2sqrt2)V'` `=2xx1.414xx40xx10^(-3)` `=113.14xx10^(-3) m^3` Similarly final pressure is given by process equation as `P'V'^(gamma)=P''V''^(gamma)` or `P''=P'((V')/(V''))^(gamma)` `=2.5xx10^5xx((40xx10^(-3))/(113.14xx10^(-3)))^(5//3)` `=4.42xx10^4` Pa |
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