# The change in internal energy of a gas $\;du=$

$(a)\;\large\frac{nR}{r-1}\;dT\qquad(b)\;nC_{v}dT\qquad(c)\;\large\frac{nC_{p}}{r} dT\qquad(d)\;All\;the\;above$

Answer : $\;All\;the\;above$
Explanation :
$du=nC_{v}dT$
$\large\frac{C_{p}}{C_{v}}=r$
$C_{v}=\large\frac{k}{r-1}$