Two different isotherms representing the rela | vedclass

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

Question

(C) The ideal gas equation is given by $pV = nRT = \frac{m}{M}RT$,where $m$ is the mass of the gas and $M$ is the molar mass.For a fixed temperature $

Vedclass · Accepted Answer

$m_1 < m_2$

Vedclass · Answer

(C) The ideal gas equation is given by $pV = nRT = \frac{m}{M}RT$,where $m$ is the mass of the gas and $M$ is the molar mass.For a fixed temperature $T$,we have $pV = 	ext{constant} 	imes m$,which implies $m = \frac{pVM}{RT}$.For a constant pressure $p$,we can see from the graph that for the same pressure,the volume $V_2$ corresponding to mass $m_2$ is greater than the volume $V_1$ corresponding to mass $m_1$ (i.e.,$V_2 > V_1$).Since $m = \frac{pVM}{RT}$,for a constant $p, T,$ and $M$,we have $m \propto V$.Since $V_2 > V_1$,it follows that $m_2 > m_1$ or $m_1 < m_2$.

Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature for the same ideal gas are shown for masses $m_1$ and $m_2$ of the gas respectively in the figure given. Then:

Similar Questions

State Boyle's law and Charles's law.

What is Boltzmann's constant?

$A$ 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 . . . . . . ${ }^{\circ} {C}$.

If the density of the Sun is $1.4 \, g \, cm^{-3}$, the pressure is $1.4 \times 10^9 \, \text{atm}$, and the molar mass of the gas in the Sun is $2 \, g \, mol^{-1}$, then the temperature of the Sun will be: [Take $R = 8.4 \, J \, mol^{-1} \, K^{-1}$]

$A$ container $A$ contains an ideal gas at pressure $P$,volume $V$,and temperature $T$. $A$ second container $B$ contains the same gas at pressure $2P$,volume $2V$,and temperature $\frac{T}{2}$. The ratio of the mass of gas in $A$ to that in $B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module