The relationship between pressure and the den | vedclass

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

Question

(C) The relationship is given by $P = K\rho$.From the ideal gas law,we know that $PV = nRT = \frac{m}{M_0} RT$,where $m$ is the mass and $M_0$ is the

Vedclass · Accepted Answer

Only if the temperature is kept constant

Vedclass · Answer

(C) The relationship is given by $P = K\rho$.From the ideal gas law,we know that $PV = nRT = \frac{m}{M_0} RT$,where $m$ is the mass and $M_0$ is the molar mass.Rearranging for density $\rho = \frac{m}{V}$,we get $P = \frac{\rho RT}{M_0}$.Comparing this with $P = K\rho$,we identify $K = \frac{RT}{M_0}$.Since $R$ is the universal gas constant and $M_0$ is the molar mass of the gas,$K$ remains constant only if the temperature $T$ is kept constant.Therefore,Boyle's law holds true only if the temperature is constant.

The relationship between pressure and the density of a gas expressed by Boyle's law,$P = K\rho$,holds true:

Similar Questions

$O_2$ gas is filled in a cylinder. If the pressure is doubled and the temperature is quadrupled,what will be the factor by which the density changes?

The volume occupied by the molecules contained in $4.5 \, kg$ of water at $STP$,if the intermolecular forces vanish away,is ........ $m^{3}$.

The gas equation $\frac{PV}{T} = \text{constant}$ is true for a constant mass of an ideal gas undergoing:

$A$ container of fixed volume contains a gas at $27^{\circ} C$. To double the pressure of the gas,the temperature of the gas should be raised to . . . . . . ${ }^{\circ} {C}$.

One mole of a gas at a pressure $2 \ Pa$ and temperature $27^{\circ} C$ is heated until both pressure and volume are doubled. What is the final temperature of the gas (in $K$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module