# 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 =  5/3) Option 1) 64P Option 2) 32P Option 3) P/64 Option 4) 16P

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

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