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In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T Delta X,where T is temperature of the system andDelta Tis the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas x = 3/2 "R h" (T/T_A)"+Rh)(V/V_A).Here, R is gas constant, V is volume of gas, T_Aand T_A are constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities. If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with P_0 V_0 = 1/3, RT_0 , the correct match is : |
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Answer» `I to Q, II toR, III toS , IV to U` `2 to 3`Isochoric process `W_(1 to 2 to 3) = W_(1to2) =P_0 (2V_0 -V_0) = P_0 V_0 =(RT_0)/3` `U_(1 to 2 to 3) = U_(1 to 2 ) +U_(2 to 3)` ` = 3/2 [P_2 V_2 - P_1 V_1 ] +3/2[P_3 V_3- P_2 V_2] =3/2 P_0 [2 V_0 - V_0]+3/2 [(3p_0)/2 xx2V_0 -p_0(2 V_0)]` `=3/2 P_0 V_0 + 3/2 P_0 V_0 = 3 P_0 V_0 = 3xx(RT_0)/3 = RT_0` `Q_(1 to 2 to 3) = Q_(1to 2)+ Q_(2 to 3)` `W_(1 to 2) + Delta u_(1 to2 ) + Delta u_(2 to 3) P_0 V_0 +3/2 P_0 V_0 + 3/2 P_0 V_0 = 4 P_0 V_0 = (4 RT_0)/3` `Q_(1 to 2)= 5/2 P_0 V_0 = 5/6 RT_0` Correct option is (B) |
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