A very tall vertical cylinder is filled with | vedclass

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

Question

(D) $1$. In a vertical column of gas,the pressure $p$ decreases as height $h$ increases due to the weight of the gas column above it. Thus,$\frac{dp}{

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) $1$. In a vertical column of gas,the pressure $p$ decreases as height $h$ increases due to the weight of the gas column above it. Thus,$\frac{dp}{dh} = -\rho g$ is the correct hydrostatic equilibrium equation for a gas.$2$. From the ideal gas law,$pV = nRT$,we can write $p = \frac{n}{V} RT$. Since the number of moles $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass),we have $p = \frac{m}{VM} RT$. Given that density $\rho = \frac{m}{V}$,we get $p = \rho \frac{RT}{M}$.$3$. Since pressure $p$ depends on the density $\rho$ and height $h$,and the weight of the gas column acts downwards,the pressure must decrease as we move up the cylinder.$4$. Therefore,all the given statements are correct.

$A$ very tall vertical cylinder is filled with a gas of molar mass $M$ under isothermal conditions at temperature $T$. The density and pressure of the gas at the base of the container are $\rho_0$ and $p_0$,respectively. Choose the correct statement$(s)$.

Similar Questions

Two thermally insulated vessels $1$ and $2$ contain air at temperatures $(T_1, T_2)$,volumes $(V_1, V_2)$,and pressures $(P_1, P_2)$. If the valve connecting them is opened,what will be the equilibrium temperature of the system?

$125 \, ml$ of gas $A$ at $0.60 \, atm$ and $150 \, ml$ of gas $B$ at $0.80 \, atm$ pressure at the same temperature are filled in a vessel of $1 \, L$ volume. What will be the total pressure of the mixture at the same temperature in $atm$?

$STATEMENT-1$: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is $1.5$ times the product of its pressure and its volume. Because
$STATEMENT-2$: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

The average translational kinetic energy of ${N}_{2}$ gas molecules at $... {}^{\circ} {C}$ becomes equal to the ${K.E.}$ of an electron accelerated from rest through a potential difference of $0.1 \ V$. (Given ${k}_{B} = 1.38 \times 10^{-23} \ J/K$). (Fill the nearest integer).

Vedclass Test Series

Exam Paper Generator

Online Exam Module