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In a thermodynamic process on an ideal m...

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 and `Delta T` is 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_A`and `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,

A

`I to P , II to T, III to Q , IV to T`

B

`I to P , II to R, III to T , IV to S`

C

`I to S, II to T, III to Q , IV to U`

D

`I to P , II to R, III to T , IV to P`

Text Solution

Verified by Experts

The correct Answer is:
D
`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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