JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 11)

A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm3 to 150 cm3 adiabatically. In this process, the value of workdone on the gas will be :
79 J
405 J
4050 J
9590 J

Explanation

For monoatomic gas degree of freedom f = 3 and $$\gamma$$ = $${5 \over 3}$$

Here for gas,

Initial pressure (P1) = 75 kPa

Initial volume (V1) = 1200 cm3

Final volume (V2) = 150 cm3

Final pressure (P2) = ?

For adiabatic process,

$${P_1}V_1^\gamma = {P_2}V_2^\gamma $$

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

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

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

$$ = 75 \times 32$$ kPa

= 2400 kPa

Work done in adiabatic process,

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

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

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

$$ = {{ - 810000 \times {{10}^{ - 3}}} \over 2}$$

$$ = - 405$$ kJ

Comments (0)

Advertisement