Three processes compose a thermodynamic cycle shown in the accompanying `P-V`, diagram of an ideal gas.
Process `1 rarr 2` take place at constant temperature, during this process `60 J` of heat enters the system.
Process `2 rarr 3` takes place at constant volume. During this process `40 J` of heat leaves the system.
Process `3 rarr 1` is adiabatic.
What is the change in internal energy of the during process `3 rarr 1`?

Options

A. `-40 J`
B. `-20 J`
C. `+ 20 J`
D. `+ 40 J`

Answer

d

Explanation

Here, process `1 to 2` : isothermal , `Delta U_(12) = 0 " and process " 2 to 3 : " isochoric, " Delta W = 0 `
`rArr Delta Q_(23) = Delta U_(23) rArr - 40 = Delta U_(23)`
Now, for a cyclic process, `Delta U = 0 `
Therefore , `Delta U_(12) + Delta U_(23) + Delta U_(31) = 0 `
`rArr 0 + (-40) + Delta U_(31) =0 `
`rArr 0 + (-40) + Delta U_(31) =0 `
`rArr Delta U_(31) = + 40 J`