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An ideal gas at 75 cm mercury pressure is compressed isothermally until its volume is reduced to three quarters of its original volume. It is then allowed to expand adiabatically to a volume 20% greater than its original volume. If the initial temperature of the gas is 17^@C,calculate the final pressure and temperature (gamma= 1.5). |
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Answer» Solution :First of all the gas is compressed isothermally. Using Boyle'slaw `P_1V_1=P_2V_2` or `P_2=(P_1V_1//V_2)` Here `P_1`=75 cm of mercury and `V_2=3/4V_1` Thus `P_2=(75V_1)/((3//4)V_1)`=100 cmof mercury The gas is now expanded ADIABATICALLY to 20% greater of its ORIGINAL value. Under adiabatic change the pressure and volume of gas are related as `P_2V_2^gamma=P_3V_3^gamma` or `P_3=P_2(V_2/V_3)^(gamma)` Here `V_2=3/4V_1` and `V_3=120/100V_1` Thus `P_3=100xx((3V_1)/4)^(1.5)XX(100/(120V_1))^(1.5)` `=100xx(3/4)^(1.5)xx(5/6)^(1.5)` `=100xx(5/8)^(1.5)` =100 x 0.494 = 49.4 cm of mercury Let the final TEMPERATURE after adiabatic change be then from the relation of temperature and volume in an adiabatic process,we have Now `T_2V_2^(gamma-1)=T_3V_3^(gamma-1)` `T_2=17^@C=(273+27)`=290 K Now `T_3=T_2(V_2/V_3)^(gamma-1)` `=290xx((3V_1)/4)^(1.5-1)xx(100/(120V_1))^(1.5-1)` `=290xx(5/8)^(0.5)`=229.3 K Hence the finaltemperaturewill be `-43.7^@C` |
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