A cylinder of capacity 200 L is filled with | vedclass

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

Question

(D) The volume of the cylinder is $V = 200 \ L = 200 	imes 10^{-3} \ m^3 = 0.2 \ m^3$.The total translational kinetic energy $E_k$ is given as $1.52

Vedclass · Accepted Answer

$5 	imes 10^5$

Vedclass · Answer

(D) The volume of the cylinder is $V = 200 \ L = 200 	imes 10^{-3} \ m^3 = 0.2 \ m^3$.The total translational kinetic energy $E_k$ is given as $1.52 	imes 10^5 \ J$.According to the kinetic theory of gases,the pressure $P$ is related to the translational kinetic energy per unit volume by the formula $P = \frac{2}{3} \frac{E_k}{V}$.Substituting the values into the formula:$P = \frac{2}{3} 	imes \frac{1.52 	imes 10^5}{0.2}$.$P = \frac{2}{3} 	imes 7.6 	imes 10^5$.$P \approx 5.06 	imes 10^5 \ N \ m^{-2}$.Rounding to the nearest given option,the pressure is $5 	imes 10^5 \ N \ m^{-2}$.

$A$ cylinder of capacity $200 \ L$ is filled with $H_2$ gas. The total average translational kinetic energy of the molecules is $1.52 \times 10^5 \ J$. Calculate the pressure of $H_2$ in the cylinder in $N \ m^{-2}$.

Similar Questions

$A$ gas is filled in a container at pressure ${P_0}$. If the mass of all molecules is halved and their $rms$ speed is doubled,what will be the new pressure?

The translational kinetic energy of the molecules of a gas at absolute temperature $T$ can be doubled by:

The total momentum of the molecules of $1 \,gm$ $mol$ of a gas in a container at rest at $300 \,K$ is

The pressure of an ideal gas is related to energy by the equation $E = \frac{3}{2} PV$. Here,$E$ stands for:

In an ideal gas at temperature $T,$ the average force that a molecule applies on the walls of a closed container depends on $T$ as $T^q$. $A$ good estimate for $q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module