Question

The K.E of mole of an ideal gas is $E=\frac{3}{2}\text{RT}$. Then $C_p$ will be

• A. $0.5\text{R}$
• B. $0.1\text{R}$
• C. $1.5\text{R}$
• D. $2.5\text{R}$

Answer 1

The correct option is D $2.5\text{R}$

$U=PE+KE$

There won't be any potential energy as we do not consider force of interaction.

$U=\frac{f}{2}1RTf=3$

$C_V=\frac{f}{2}R{C}_p=(\frac{f}{2}+1)R$

$C_V=\frac{3}{2}R{C}_p=\frac{5}{2}R$