One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is…

1. The change in internal energy in whole cyclic process is 250 R
2. The change in internal energy in the process CA is 700 R.
3. The change in internal energy in the process AB is  - 350 R
4. The change in internal energy in the process BC is  - 500 R

Solution

$\Delta {\text{U}}_{\text{CA}}={\text{nc}}_{\text{v}}\left({\text{T}}_{\text{A}}-{\text{T}}_{\text{C}}\right)=1\times \frac{5}{2}\text{R}\left(-200\right)=-500\text{R}$

$\Delta {\text{U}}_{\text{AB}}={\text{nc}}_{\text{v}}\left({\text{T}}_{\text{B}}-{\text{T}}_{\text{A}}\right)=1\times \frac{5}{2}\text{R}\times 400=+1000\text{R}$

$\Delta {\text{U}}_{\text{BC}}={\text{nc}}_{\text{v}}\left({\text{T}}_{\text{C}}-{\text{T}}_{\text{B}}\right)=1\times \frac{5}{2}\text{R}\left(-200\right)=-500\text{R}$

Asked in: JEE Main 2014 (06 Apr)

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