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Two different isotherms representing the relationship between pressure \mathrm{p} and volume \mathrm{V} at a given temperature of the same ideal gas are shown for masses \mathrm{m_1 \: and \: m_2}, then


 

Option: 1

\mathrm{m_1>m_2}
 


Option: 2

\mathrm{m_1=m_2}
 


Option: 3

\mathrm{m_1<m_2}
 


Option: 4

Nothing can be predicted


Answers (1)

best_answer

\mathrm{p V=n R T=\frac{m}{M} R T}

For lst graph, \mathrm{\quad p=\frac{m_1}{M} \frac{R T}{V_1}}      .............(i)

For lind graph,\mathrm{ \quad p=\frac{m_2}{M} \frac{R T}{V_2}}         ..........(ii)

On equating Eqs. (i) and (ii), we get

\mathrm{ \frac{m_1}{m_2}=\frac{V_1}{V_2} \Rightarrow m \propto V }

\mathrm{ As, V_2>V_1 \Rightarrow m_1<m_2 }

Hence option 3 is correct.

 

Posted by

Sanket Gandhi

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