Question

When an ideal gas is taken from state `a` to `b`, along a path `acb, 84 kJ` of heat flows into the gas and the gas does `32 kJ` of work. The following conclusions are drawn. Mark the one which is not correct.

A. If the work done along the path `adb` is `10.5 kJ`, the heat will flow into the gas is `62.5 kJ`
B. When the gas is returned from `b` to `a` along the curved path, the work done on the gas is `21 kJ`, and the system absorbs `73 kJ` of heat
C. If `U_(a) = 0, U_(d) = 42 kJ`, and the work done along the path `adb` is `10.5 kJ` then the heat absorbed in the process `ad` is `52.4 kJ`.
D. If `U _(a) = 0, U_(d) = 42 kJ`, heat absorbed in the precess `db` is `10 kJ`.

Answer 1

Correct Answer - B
b. For path `acb`:
`Delta Q = Delta U + Delta W`
`implies 84 = Delta U + 32 implies Delta U = 52 kJ`
Hence `Delta U_(acb) = Delta U_(ab) = Delta U_(adb) = 52 J`
For path `adb`:
`Delta Q = Delta U + Delta W`
`= 52 + 10.5 = 52.5 kJ`
So option (a) is correct.
For process `ba`, system will release the heat. So option (b) is wrong.
For path `ad`:
`Delta W_(adb) = Delta W_(ad) + Delta W_(db)`
`implies 10.5 = Delta W_(ad) + 0`
`implies Delta W_(ad) = 10.5 kJ`
`Delta Q_(ad) = Delta U_(ad) + Delta W_(ad)`
`= (42 - 0) + 10.5`
`= 52.5 kJ`
So option (c ) is correct.
`Delta Q_(adb) = 52 + 10.5 = 62.5 kJ`
`Delta Q_(db) = Delta Q_(adb) - Delta Q_(ad)`
`= 62.5 52.5`
`= 10 kJ`
So option (d) is correct.
Hence answer of this question is (b)