Two different isotherms representing the relationship between pressure p and volume V at a given temperature of the same ideal gas are shown for masses m1 and m2 of the gas respectively in the figure given, then

Question

Two different isotherms representing the relationship between pressure p and volume V at a given temperature of the same ideal gas are shown for masses m1 and m2 of the gas respectively in the figure given, then (A) m1 > m2 (B) m1=m2 (C) m1< m2 (D) m1 ≥ m2

Tardigrade Admin · Accepted Answer

p V=n R T=(m/M) R T

For 1 textst graph, p=(m1/M) (R T/V1) For 2 text nd graph, p=(m2/M) (R T/V2) Equating the two, we get (m1/m2)=(V1/V2) ⇒ m propto (1/V) As V2 > V1 ⇒ m1 < m2

Q. Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ of the gas respectively in the figure given, then

$m_{1} > m_{2}$

$m_{1} = m_{2}$

$m_{1} < m_{2}$

$m_{1} \geq m_{2}$

Q. Two different isotherms representing the relationship between pressure p and volume V at a given temperature of the same ideal gas are shown for masses m1​ and m2​ of the gas respectively in the figure given, then

m1​>m2​

m1​=m2​

m1​<m2​

m1​≥m2​

pV=nRT=Mm​RT For 1st graph, p=Mm1​​V1​RT​ For 2 nd graph, p=Mm2​​V2​RT​ Equating the two, we get m2​m1​​=V2​V1​​ ⇒m∝V1​ As V2​>V1​ ⇒m1​<m2​

Q. Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ of the gas respectively in the figure given, then

$m_{1} > m_{2}$

$m_{1} = m_{2}$

$m_{1} < m_{2}$

$m_{1} \geq m_{2}$