An ideal gas undergoes a thermodynamic cycle as shown in Fig. Which of the following statement are correct ?

Options

A. Straight line `AB` cannot pass through `O`.
B. During process `AB`, temperature decreases while during process `BC` it increases.
C. During process `BC`, work is done by the gas against external pressure and temperature of the gas increases.
D. During process `CA`, work is done by the gas against external pressure and heat supplied to the gas is exactly equal to this work.

Answer

B, D

Explanation

For an isothermal process, `PV=` constant
`P=( consta nt )/(V) ` or `P prop (1)/(V)`
It means, for an isothermal process the graph between `P` and `1//V` will be straight line passing through origin. Hence, the straight line `AB` will pass through origin. Hence, option `(a)` is wrong.
During process `AB` , the pressure `P` remains constant but `1//V` increases. It means, volume `V` decreases. Hence, `AB` is isobaric compression. Since volume of the gas decreases at constant pressure, therefore its temperature decreases. But temperature at `C` is equal to that of `A`. Hence during the process `BC`, the temperature of the gas increases. In face, process `BC` is an isobhoric heating. Therefore, option `(b)` is correct. Since the during the process `BC`, volume remains constant, no work is done by the gas against external pressure. Therefore option `(c)` is wrong.
During process `CA, 1//V` decreases, it means volume `V` increases. Since the volume increases, work is done by the gas against external pressure.
Since process `CA` is an isothermal process, no chage ini internal energy of the gas takes place. Hence, according to the first law of thermodynamics, `Q=W+ DeltaU`, heat suplied during this process is equal to work done by the gas against external pressure. Hence, option `(d)` is correct.