When text500 calories heat is given to the gas textX in an isobaric process, its work done comes out as text142 text.8 calories. The gas textX is

Question

When text500 calories heat is given to the gas textX in an isobaric process, its work done comes out as text142 text.8 calories. The gas textX is (A)  textO text2 (B)  textN textH3 (C)  textH e (D)  textS textO2

Tardigrade Admin · Accepted Answer

∵ Δ T =( W / nR ) ∴ Q = nC p (Δ T )= nC p ( W / nR )=( C p W / R ) Cp=(Q R/W)=(500 × 2/142.8)=7 C p =7, indicates that the gas is diatomic. Thus, it should be O 2

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

$O_{2}$

$N H_{3}$

$H e$

$S O_{2}$

$∵ Δ T = \frac{W}{n R}$

$∴ Q = n C_{p} (Δ T) = n C_{p} \frac{W}{n R} = \frac{C _{p} W}{R}$

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

$C_{p} = 7,$ indicates that the gas is diatomic. Thus, it should be $O_{2}$

Q. When 500 calories heat is given to the gas X in an isobaric process, its work done comes out as 142.8 calories. The gas X is

O2​

NH3​

He

SO2​