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Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure. (a) Find the work done when the gas is taken from state 1 to state 2. (b) What is the ratio of temperature T_1/T_2 if V_2=2V_1 ? (c) Given the internal energy for one mole of gas at temperature T is 3/2RT, find the heat 2 supplied to the gas when it is taken from state 1 to 2, with V_2 = 2V_1 |
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Answer» Solution :From graph, `PV^(1/2)`=constant K `therefore P=K/sqrtV` (a)WORK DONE for the process 1 to 2, `DeltaW=int_(V_1)^(V_2)P dV = int_(V_1)^(V_2)K/sqrtV xx dV` `DeltaW=K[V^(1/2)/(1/2)]_(V_1)^(V_2) =2K[V_2^(1/2)-V_1^(1/2)]`…(1) but `P_1V_1^(1/2)=K` and `P_2V_2^(1/2)=K` `therefore W=2P_1V_1^(1/2)[V_2^(1/2)-V_1^(1/2)]` ....(2) (b) From ideal gas equation of state , PV=nRT `therfore T=(PV)/(nR)` `therefore T=(PsqrtV.sqrtV)/(nR)` `therefore T=(KsqrtV)/(nR)[ because PsqrtV=K]` `therefore T_1=(KsqrtV_1)/(nR)` and `T_2=(KsqrtV_2)/(nR)` `therefore T_1/T_2=sqrt(V_1/V_2)=sqrt((V_1)/(2V_1))[ because V_2=2V_1]`....(3) `therefore T_1/T_2=1/sqrt2 therefore T_2=sqrt2T_1`...(4) (c) Internal energy of gas , `U=3/2 RT` `therefore DeltaU=U_2-U_1=3/2R(T_2-T_1)` `DeltaU=3/2R(sqrt2T_1-T_1)=3/2RT_1(sqrt2-1)`....(5) `rArr DeltaW=2P_1V_1^(1/2)[sqrtV_2- sqrtV_1]` (From equation 2) `=2P_1V_1^(1/2)[sqrt(2V_1)-sqrtV_1]` `=2P_1V_1[sqrt2-1]` `=2RT_1[sqrt2-1][ because P_1V_1=RT_1]` ....(6) From first rule of thermodynamics , `DeltaQ=DeltaU+DeltaW` `=3/2RT_1(sqrt2-1)+2RT_1(sqrt2-1)` `therefore DeltaQ=(7RT_1)/2 (sqrt2-1)` |
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