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One mole of a gas is isothermally expanded at 27^@C till the volume is doubled. Then it is adiabatically compressed to its original volume. Find the total workdone. (gamma= 1.4 andR = 8.4 joule/mole/.^@K). |
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Answer» Solution :In case of isothermal expansion, the workdone is given by `dW_1=RT.ln(V_f//V_i)` HereR=8.314, `T=27^@C` =273+27=`300^@K`, `(V_f//V_i)`=2 or `dW_1=8.314xx300xxln(2)` =8.314 x 300 x 0.693 =1728.48 J. Now the gas is adiabatically COMPRESSED to its original volume. Initially at the beginning of adiabatic COMPRESSION, the TEMPERATURE of the gas is 300 K and at the end of adiabatic compression, the temperature becomes `T_2` because the temperature is changed. The INITIAL volume of the gas is `2V_1` and after compression it again becomes the original volume i.e.,`V_i` .. For an adiabatic process, we know temperature and volume are RELATED as `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` or `300xx(2V_i)^(gamma-1)=T_2xx(V_i)^(gamma-1)` or `300xx(2)^(1.4-1) =T_2(1)^(1.4-1)` or `T_2=300xx(2)^(0.4)`=395.85 K Workdone during adiabatic process is `W_2=(nR)/(gamma-1)(T_2-T_1)` `=8.314/(-(1.4-1))(395.85-300)` =-1992.24 J Total workdone=-1728.47-1992.24 =-263.77 J |
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