Question

As a result of isobaric heating by $\Delta T=72K$, one mole of certain ideal gas obtains an amount of heat $Q=1.60kJ$. The value of $\gamma$ is

• A. $1.60$
• B. $1.33$
• C. $1$
• D. $0.66$

Answer 1

The correct option is A $1.60$

For isobaric process, heat absorbed $\Delta Q=n{C}_p\Delta T$
where, n = no. of moles = 1,
$\Rightarrow C_p=\frac{1600}{72}\Rightarrow \gamma =\frac{C_p}{C_p-R}=1.6$