Indicate the correct statement for a 1 L sam | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The average translational kinetic energy $(KE)$ per molecule is given by $\frac{3}{2}kT$,which depends only on temperature. Since both gases are a

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) The average translational kinetic energy $(KE)$ per molecule is given by $\frac{3}{2}kT$,which depends only on temperature. Since both gases are at $298 \ K$,their average translational $KE$ per molecule is the same.According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain the same number of molecules $(n)$. Since $KE_{total} = n 	imes (	ext{average } KE 	ext{ per molecule})$,the total translational $KE$ is also the same.Density $(d)$ is given by $d = \frac{PM}{RT}$. Since $P$,$T$,and $R$ are constant,$d \propto M$. The molar mass of $N_2$ $(28 \ g/mol)$ is less than that of $CO_2$ $(44 \ g/mol)$,so the density of $N_2$ is less than that of $CO_2$.Therefore,all statements are correct.

Indicate the correct statement for a $1 \ L$ sample of $N_{2(g)}$ and $CO_{2(g)}$ at $298 \ K$ and $1 \ atm$ pressure.

Similar Questions

Assertion : Wet air is heavier than dry air.
Reason : The density of dry air is more than density of water.

Which one of the following is the wrong statement about the liquid?

Which one of the following statements is not correct?

Two thin circular discs $A$ and $B$ of radii $2 \ cm$ and $4 \ cm$ are in a liquid at the same depth. $T_A$ is the thrust on $A$ and $T_B$ is the thrust on $B$. Then $T_A : T_B =$

At constant temperature,$200 \ cm^3$ of $N_2$ at $720 \ mm$ pressure and $400 \ cm^3$ of $O_2$ at $750 \ mm$ pressure are taken in a $1 \ L$ flask. What will be the total pressure of the mixture in $mm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module