Test Index

Thermodynamics Part 2

Section: Physics

© examsnet.com

Question : 17 of 100

Marks: +1, -0

Thermodynamic process is shown below on a

p

-V diagram for one mole of an ideal gas.

V_{2}=2 V_{1}

, then the ratio of temperature

\;\frac{T_{2}}{T_{1}}

[25 Feb 2021 Shift 2]

$\;\frac{1}{\sqrt{2}}$
$\sqrt{2}$
$\;\frac{1}{2}$
2

Solution:

Here,

p_{1}

and

p_{2}, T_{1}

and

T_{2}, V_{1}

and

V_{2}

are initial and final pressures, temperatures and volumes, respectively.

Given,

\;\; V_{2}=2 V_{1}

p V^{1 / 2} = \;\text{ constant }\;

From graph,

\gamma=

adiabatic constant

=\frac{1}{2}

\Rightarrow \;\; p_{1} V_{1}^{1/2}=p_{2} V_{2}^{1/2} \Rightarrow \;\frac{p_{1}}{p_{2}}=\left(\;\frac{2 V_{1}}{V_{1}}\right)^{1/2}=2^{1/2}

Also,

\;\; p^{1-\gamma} T^{\gamma}=

constant (for adiabatic process)

\Rightarrow \;\; p_{1}^{1-\gamma} T_{1}^{\gamma}=p_{2}^{1-\gamma} T_{2}^{\gamma}

\Rightarrow \;\; \left(\;\frac{p_{1}}{p_{2}}\right)^{1-\gamma} = \left(\;\frac{T_{2}}{T_{1}}\right)^{\gamma} \Rightarrow \left(\;\frac{p_{1}}{p_{2}}\right)^{\frac{1-\gamma}{\gamma}} = \;\frac{T_{2}}{T_{1}}

\therefore \;\; \frac{T_{2}}{T_{1}} = (2^{1/2})^{\frac{1-0.5}{0.5}} = \sqrt{2}

© examsnet.com

Go to Question:

Prev Question

Next Question