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Please help! - Thermodynamics - NEET

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. the final pressure of the gas is: 

take\:\:\left( \gamma= \frac{5}{3} \right ) 

  • Option 1)

    64 P

  • Option 2)

    32 P

  • Option 3)

    \frac{P}{64}

  • Option 4)

    16 P

 
Answers (1)
83 Views

As learnt in

Equation of state -

dQ= 0

n, C_{V}, dT+PdV= 0
 

- wherein

On solving

gamma frac{dV}{V}+frac{dP}{P}= 0

Rightarrow PV^{gamma }= constant

 

 For Isothermal expansion,

P_{i}= P, \ V_{i}= V,\ V_{1}= 2V\ \ \Rightarrow P_{1} =P_{i}\cdot \frac{V}{2V}= \frac{P}{2}

For adiabatic expansion,

P_{1}= \frac{P}{2},\ V_{1}=2V,\ V_{f}= 16V,\ P_{f} =?

PV^{\gamma }= \ constant

\Rightarrow P_{1}V_{1}^{\gamma } = P_{f}V_{f}^{\gamma }\ \ \Rightarrow P_{f}= P_{1}\cdot \left ( \frac{V_{1}}{V_{f}} \right )^{\gamma }

P_{f}= \frac{P}{2}\cdot \left ( \frac{2V}{16V} \right )^{\frac{5}{3}} = \frac{P}{2}\cdot \left ( \frac{1}{2} \right )^{5} = \frac{P}{64} 


Option 1)

64 P

This option is incorrect

Option 2)

32 P

This option is incorrect

Option 3)

\frac{P}{64}

This option is correct

Option 4)

16 P

This option is incorrect

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