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A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is:

Option: 1

\frac{13\; R}{6}         


Option: 2

\frac{19\; R}{6}


Option: 3

\frac{23\; R}{6}


Option: 4

\frac{17\; R}{6}


Answers (1)

best_answer

 Let initial pressure, volume, temperature be P_{0},\; V_{0},\; T_{0} indicated by state A in P-V diagram.

The gas is then isochorically taken to state B\left ( 2P_{0},\; V_{0},\; 2T_{0} \right ) and then taken from state B  to state C\left ( 2P_{0},\; 2V_{0},\; 4T_{0} \right )  isobarically.

`Total heat absorbed by 1 mole of gas

\Delta Q \; \; =C_v\left ( 2T_{0}-T_{0} \right )+C_{P}\left ( 4T_{0}-2T_{0} \right )

  =   \frac{5}{2}RT_{0}+\frac{7}{2}R\times 2T_{0}=\frac{19}{2}RT_{0}

Total change in temperature from state A to C is      

\Delta T=3T_{0}

              

\text{Average molar heat capacity}=\frac{\Delta Q}{\Delta T}=\frac{\frac{19}{2}RT_{0}}{3T_{0}}=\frac{19}{6}R.

 

 

 

 

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Nehul

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