One kg of a diatomic gas is at a pressure of | vedclass

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

Question

(D) For a diatomic gas,the internal energy $U$ is given by $U = \frac{f}{2} PV$,where $f$ is the degrees of freedom.For a diatomic gas,$f = 5$.Thus,$U

Vedclass · Accepted Answer

$5 	imes 10^4 \ J$

Vedclass · Answer

(D) For a diatomic gas,the internal energy $U$ is given by $U = \frac{f}{2} PV$,where $f$ is the degrees of freedom.For a diatomic gas,$f = 5$.Thus,$U = \frac{5}{2} PV$.Given:Pressure $P = 8 	imes 10^4 \ N/m^2$.Mass $m = 1 \ kg$.Density $ho = 4 \ kg/m^3$.Volume $V = \frac{m}{ho} = \frac{1 \ kg}{4 \ kg/m^3} = 0.25 \ m^3$.Substituting the values into the formula:$U = \frac{5}{2} 	imes (8 	imes 10^4 \ N/m^2) 	imes (0.25 \ m^3)$.$U = 2.5 	imes 8 	imes 10^4 	imes 0.25$.$U = 20 	imes 10^4 	imes 0.25 = 5 	imes 10^4 \ J$.

One kg of a diatomic gas is at a pressure of $8 \times 10^4 \ N/m^2$. The density of the gas is $4 \ kg/m^3$. What is the energy of the gas due to its thermal motion?

Similar Questions

An ideal gas is kept in a cylinder of volume $3 \,m^3$ at a pressure of $3 \times 10^5 \,Pa$. The energy of the gas is

The kinetic energy of $1 \ g$ of $He$ gas at $272^{\circ}C$ is ........... $J$.

The average force applied on the walls of a closed container depends on temperature $(T)$ as ($T$ is the temperature of an ideal gas).

The temperature at which the kinetic energy of oxygen molecules becomes double its value at $27^{\circ}\,C$ is $............^{\circ}\,C$.

The pressure of a gas is $6 \times 10^4 \ N/m^2$. What will be the kinetic energy per unit volume of the gas molecules?

Vedclass Test Series

Exam Paper Generator

Online Exam Module