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Thermodynamics Part 3

Section: Physics

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Question : 24 of 100

Marks: +1, -0

A sample of monoatomic gas is taken at initial pressure of

75\ \mathrm{kPa}

. The volume of the gas is then compressed from 1200

\mathrm{cm}^3

150\ \mathrm{cm}^3

adiabatically. In this process, the value of workdone on the gas will be :

[30-Jun-2022-Shift-1]

$79\ \mathrm{J}$
$405\ \mathrm{J}$
$4050\ \mathrm{J}$
$9590\ \mathrm{J}$

Solution:

For monoatomic gas degree of freedom

f=3

and

\gamma=\frac{5}{3}

Here for gas,

Initial pressure

(P_1)=75\ \mathrm{kPa}

Initial volume

(V_1)=1200\ \mathrm{cm}^3

Final volume

(V_2)=150\ \mathrm{cm}^3

Final pressure

(P_2)=

?

For adiabatic process,

P_1 V_1^\gamma = P_2 V_2^\gamma

\Rightarrow 75 \times (1200)^\gamma = P_2 (150)^\gamma

\Rightarrow P_2 = 75 \times \left(\frac{1200}{150}\right)^{\frac{5}{3}}

= 75 \times (8)^{\frac{5}{3}}

= 75 \times 32\ \mathrm{kPa}

= 2400\ \mathrm{kPa}

Work done in adiabatic process,

W = \frac{P_2 V_2 - P_1 V_1}{1-\gamma}

= \frac{2400 \times 10^3 \times 150 \times 10^{-6} - 75 \times 10^3 \times 1200 \times 10^{-6}}{1 - \frac{5}{3}}

= \frac{(2400 \times 150 - 75 \times 1200) \times 10^{-3}}{-\frac{2}{3}}

= \frac{-810000 \times 10^{-3}}{2}

= -405\ \mathrm{kJ}

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