Question

When $\text{500}$ calories heat is given to the gas $\text{X}$ in an isobaric process, its work done comes out as $\text{142} \text{.8}$ calories. The gas $\text{X}$ is

• A. $\text{O}_{\text{2}}$
• B. $\text{N} \text{H}_{3}$
• C. $\text{H} e$
• D. $\text{S}\text{O}_{2}$

Answer 1

Answer: (A)

$\because \Delta T =\frac{ W }{ nR }$

$\therefore Q = nC _{ p }(\Delta T )= nC _{ p } \frac{ W }{ nR }=\frac{ C _{ p } W }{ R }$

$C_{p}=\frac{Q R}{W}=\frac{500 \times 2}{142.8}=7$

$C _{ p }=7,$ indicates that the gas is diatomic. Thus, it should be $O _{2}$