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

$N_2$ gas at $100 \, kPa$ pressure is filled in a $1 \, L$ flask and $O_2$ gas at $320 \, kPa$ pressure is filled in another $3 \, L$ flask. If both flasks are connected,what will be the resultant pressure in $kPa$?

The mass of a gas in a container at $1 \, \text{atm}$ pressure and $200 \, K$ temperature is $3.2 \, g$. What will be the mass of the same gas in the same container at $500 \, K$ temperature and $10 \, \text{atm}$ pressure?

The density of a gas at $27\,^oC$ and $1\, atm$ is $d$. At constant pressure,at which of the following temperatures will its density become $0.75\, d$?

At what $^o C$ temperature,the sample of neon gas would be heated to double its pressure,if the initial volume of gas is reduced to $15\%$ at ${75\,^o}C$?

According to Charles's Law,what is the slope of the graph of volume versus temperature? ($V_0$ is the volume of the gas at $0\,^{\circ}C$ and $V_t$ is the volume of the gas at $t\,^{\circ}C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module