Which amongst the given plots is the correct | vedclass

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

Question

(B) For an ideal gas,the ideal gas equation is $pV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $pV = \frac

Vedclass · Accepted Answer

Vedclass · Answer

(B) For an ideal gas,the ideal gas equation is $pV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $pV = \frac{m}{M} RT$. Rearranging this gives $p = \left( \frac{RT}{M} \right) \left( \frac{m}{V} \right)$. Since density $d = \frac{m}{V}$,the equation becomes $p = \left( \frac{RT}{M} \right) d$. This is in the form of a straight line $y = mx$,where the slope is $\frac{RT}{M}$. Since the slope is directly proportional to temperature $(T)$,a higher temperature will result in a steeper slope. Given $T_3 > T_2 > T_1$,the slope order will be $T_3 > T_2 > T_1$. Therefore,the plot with the steepest line for $T_3$ and the least steep for $T_1$ is correct.

Which amongst the given plots is the correct plot for pressure $(p)$ $vs$ density $(d)$ for an ideal gas?

Similar Questions

The correct value of the gas constant $R$ is close to

The density of neon will be highest at

The compressibility factor of an ideal gas is

In the equation $PV = nRT$,which one cannot be the numerical value of $R$?

Calculate the volume occupied by $8.8 \,g$ of $CO_{2}$ at $31.1^{\circ} C$ and $1 \,bar$ pressure.
( $R=0.083 \,bar \,L \,K^{-1} \,mol^{-1}$ )

Vedclass Test Series

Exam Paper Generator

Online Exam Module