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  • Help me please, ‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be :

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure.  The maximum temperature of the gas during the process will be :

  • Option 1) (9P0V0/4nR) 
  • Option 2) (3P0V0/2nR)  
  • Option 3) (9P0V0/2nR)  
  • Option 4)(9P0V0/nR)  
 

Answers (1)

best_answer

As we learnt in

Ideal gas equation -

PV = nRT
 

 

- wherein

T= Temprature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

 

 At any point between A & B we can write relation between P & V by using equation of straight line

 V-V_{0}=\frac{2V_{0}-V_{0}}{P_{0}-2P_{0}}(P-2P_{0})

V-V_{0}=\frac{-V_{0}}{P_{0}}(P-2P_{6})

P\left(\frac{-V_{0}}{P_{0}} \right )+2V_{0}=V-V_{0}

P=\frac{-P_{0}}{V_{0}}(V-3V_{0})

From ideal gas equation 

    PV = nRT

\Rightarrow\ \;\frac{nRT}{V}=\frac{-P_{0}}{V_{0}}(V-3V_{0})

T=\frac{-P_{0}}{nRV_{0}}(V^{2}-3V_{0}V)

For temperature to be maximum at any point \frac{dT}{dV}=0

\Rightarrow\ \;2V-3V_{0}=0

    \therefore\ \; V=\frac{3V_{0}}{2}

\therefore\ \; T_{max}=\frac{-P_{0}}{nRV_{0}} \left(\frac{9}{4}V_{0}^{2}-\frac{9}{2}V_{0}^{2} \right )=-\frac{P_{0}}{nRV_{0}}.\frac{-9}{4}V_{0}^{2}=\frac{9}{4}\frac{P_{0}V_{0}}{nR}

Correct option is 1.

 

 


Option 1)

\frac{9P_{0}V_{0}}{4nR}

This is the correct option.

Option 2)

\frac{3P_{0}V_{0}}{2nR}

This is an incorrect option.

Option 3)

\frac{9P_{0}V_{0}}{2nR}

This is an incorrect option.

Option 4)

\frac{9P_{0}V_{0}}{nR}

This is an incorrect option.

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