The isotherms of an ideal gas at T_1, T_2, T_ | vedclass

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

Question

(D) For an ideal gas,the equation is $PV = nRT$,which can be rearranged as $P = (nRT) 	imes (\frac{1}{V})$.Comparing this with the equation of a stra

Vedclass · Accepted Answer

$m_1 > m_2 > m_3$

Vedclass · Answer

(D) For an ideal gas,the equation is $PV = nRT$,which can be rearranged as $P = (nRT) 	imes (\frac{1}{V})$.Comparing this with the equation of a straight line $y = mx + c$,where $y = P$,$x = \frac{1}{V}$,and $c = 0$,the slope $m$ is equal to $nRT$.Since $n$ and $R$ are constants,the slope $m$ is directly proportional to the temperature $T$ $(m \propto T)$.Given the condition $T_1 > T_2 > T_3$,it follows that the slopes must satisfy the order $m_1 > m_2 > m_3$.

The isotherms of an ideal gas at $T_1, T_2, T_3$ along with their slopes $(m)$ are shown in the graph. If $T_1 > T_2 > T_3$,then the correct order of slopes of these isotherms is:

Similar Questions

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

If the temperature of a fixed mass of gas is increased by $1\,^{\circ}C$ at constant pressure,what will be the new volume of the gas?

Which of the following does not represent Charles's Law?

For an ideal gas,which of the following graphs will not be a straight line?

The pressure and temperature of $4 \ dm^3$ of carbon dioxide gas are doubled. Then the volume of carbon dioxide gas would be ............. $dm^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module