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A monoatomic gas at pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16 V. The final pressure of the gas is : (take \gamma =  5/3)

  • Option 1)

    64P

  • Option 2)

    32P

  • Option 3)

    P/64

  • Option 4)

    16P

 

Answers (1)

best_answer

As we 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 first process

T= constant, V\rightarrow 2V

\because P _{1}V _{1}=P _{2}V _{2}.

\Rightarrow P_{2}=P_{1}.\left ( \frac{V_{1}}{V_{2}} \right )=P.\left ( \frac{V}{2V} \right )=\frac{P}{2}

For second process

PV^{r}= constant

r=\frac{5}{3}

\Rightarrow \frac{P}{2}.\left ( 2V \right )^{r}=P_{f}.\left ( 16V \right )^{r}

\Rightarrow P_{f}-\frac{P}{2}.\left ( \frac{2V}{16V} \right )^{r}=\frac{P}{2}.\left ( \frac{1}{8} \right )^{\frac{5}{3}}

=\frac{P}{2}.\left ( \frac{1}{32} \right )=\frac{P}{64}

\therefore P_{f}=\frac{P}{64}

Correct option is 3.


Option 1)

64P

Incorrect 

Option 2)

32P

Incorrect 

Option 3)

P/64

Correct 

Option 4)

16P

Incorrect 

Posted by

divya.saini

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