2305\dpi{200} \fn_cs \textit{We know that,}
\inline \dpi{200} \fn_cs C_{p}-C_{v}=R
\dpi{200} \fn_cs \Rightarrow C_{p}=C_{v}+R = 20.75 \textit{ (If you take R = 8.3)}
\dpi{200} \fn_cs \textit{Work done = }C_{p}\cdot (\bigtriangleup \theta ) = 2075
\dpi{200} \fn_cs \textit{Change in internal energy = }2075-2075 = 0J
- Shaswata Roy I got the previous answer by taking R=8.314 ( a more accurate value of R).
Akash Anand Good Work 
Soumyabrata Mondal thnkw :)