1.

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 on one mole of monatomic ideal gas is as shown in the TV-diagram with P_0 V_0 = 1/3 RT_0,the correct match is,

Answer»

`I to P , II toT, III toQ , IV to T`
`I to P , II toR, III toT , IV to S`
`I to S, II toT, III toQ , IV to U`
`I to P , II toR, III toT , IV to P`

SOLUTION :`1 to 2`ISOTHERMAL process
`2 to 3`Isochoric process
`W_(1to2) =" nRT In"V_2/V_1 = 1xxR xxT_0/3 "In"(2V_0)/V_0 = (RT_0)/3 "In 2"`
`U_(1 to 2to3) = U_(2 to 3) = 3/2 "nR"Delta T = 3/2 (1) xxR xx (2T_0)/3 = RT_0`
`Q_(1 to 2 to 3) = W_(1 to 2) +U_(2to3) =(RT_0)/3 (3 + "In 2")`
`Q_(1 to 2 to 3) = W_(1 to 2) =(RT_0)/3 "In 2"`


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