# A mass of diatomic gas $(\gamma =1.4)$ at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $27^{o}C$ to $927^{o}C$. The pressure of the gas in final state is:  Option 1) 28 atm       Option 2) 68.7 atm Option 3) 256 atm Option 4) 8 atm

P Prateek Shrivastava

As learnt in

Equation of state -

$dQ= 0$

$n\, C_{V}\, dT+PdV= 0$

- wherein

On solving

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

$\Rightarrow PV^{\gamma }= constant$

In adiabatic process, $p^{1 - \gamma }\cdot T^{\gamma }= \ constant$

$\Rightarrow {p_{i}}^{1-\gamma }\cdot {T_{i}}^{\gamma } = {p_{f}}^{1-\gamma }\cdot {T_{f}}^{\gamma }$

$\Rightarrow p_{f} = \left ( {p_{i}}^{1-\gamma } \cdot \frac{{T_{i}}^{\gamma }}{{T_{f}}^{\gamma }}\right )^{\frac{1}{1-\gamma }} = p_{i}\cdot \left ( \frac{T_{i}}{T_{f}} \right )^{\frac{\gamma }{1-\gamma }}$

$p_{i} = 2\ atm, \ \gamma = 1.4$

$T_{i} = 300\ K, \ T_{f} = 1200\ K$

$\Rightarrow p_{f} = \left ( 2\ atm \right )\left ( \frac{300}{1200} \right )^{\frac{1.4}{-0.4}} = 2\ atm\cdot \left ( 2^{7} \right )$

$p_{f} = 256\ atm$

Option 1)

28 atm

This option is incorrect

Option 2)

68.7 atm

This option is incorrect

Option 3)

256 atm

This option is correct

Option 4)

8 atm

This option is incorrect

Exams
Articles
Questions