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

In the given $(V - T)$ diagram,what is the relation between pressures $P_1$ and $P_2 \,?$

$A$ gas is contained in a $5 \, L$ vessel at a pressure of $0.8 \, m$. It is connected to another evacuated vessel of $3 \, L$ capacity. What will be the resultant pressure in $m$ (at constant temperature)?

When the pressure of a gas is increased by $0.4\%$,its temperature increases by $1\,^{\circ}\text{C}$. What is the initial temperature of the gas?

$A$ perfect gas of a given mass is heated first in a small vessel and then in a large vessel,such that their volumes remain unchanged. The $P-T$ curves are

$A$ cylinder of $5 \ L$ capacity,filled with air at $N.T.P.$,is connected to another evacuated cylinder of $30 \ L$ capacity. The resultant air pressure in both the cylinders will be ...... $cm$ of $Hg$. (in $.85$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module