For the same mass of two different ideal gase | vedclass

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

Question

(A) For an ideal gas, the ideal gas equation is $pV = \frac{w}{M}RT$, where $w$ is the mass, $M$ is the molecular weight, $R$ is the gas constant, and

Vedclass · Accepted Answer

$M_{1} > M_{2}$

Vedclass · Answer

(A) For an ideal gas, the ideal gas equation is $pV = \frac{w}{M}RT$, where $w$ is the mass, $M$ is the molecular weight, $R$ is the gas constant, and $T$ is the temperature.Since $w$, $R$, and $T$ are constant, let $wRT = K$. Then $pV = \frac{K}{M}$, or $V = \frac{K}{Mp}$.Taking the logarithm on both sides: $\log V = \log \left( \frac{K}{M} \right) - \log p$.This is in the form of $y = mx + c$, where $y = \log V$, $x = \log p$, $m = -1$, and the intercept $c = \log \left( \frac{K}{M} \right)$.From the graph, the intercept for $M_{2}$ is higher than the intercept for $M_{1}$.Therefore, $\log \left( \frac{K}{M_{2}} \right) > \log \left( \frac{K}{M_{1}} \right)$.This implies $\frac{K}{M_{2}} > \frac{K}{M_{1}}$, which means $M_{1} > M_{2}$.

For the same mass of two different ideal gases of molecular weights $M_{1}$ and $M_{2}$, plots of $\log V$ vs $\log p$ at a given constant temperature are shown. Identify the correct option.

Similar Questions

$A$ $10 \ g$ of a gas at atmospheric pressure is cooled from $273 \ ^oC$ to $0 \ ^oC$ keeping the volume constant,its pressure would become ............. $atm$.

$A$ container of volume $2.24 \, L$ can withstand a maximum pressure of $2 \, atm$ at $298 \, K$ before exploding. The maximum amount of nitrogen (in $g$) that can be safely put in this container at this temperature is closest to

If $P$,$V$,and $T$ represent the pressure,volume,and temperature of a gas respectively,which is the correct representation of Boyle's Law?

The volume-temperature graphs of a given mass of an ideal gas at different constant pressures are shown below. What is the correct order of pressures?

Densities of a gas at different conditions are in the ratio $1:2$ and their temperatures are in the ratio $2:1$,then the ratio of their respective pressure is :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module