Verified Solution • 2025–26 Session moderate

QJEEM20Phf439: JEE Main 2020 Physics - Class 11 JEEM20Phf439 Solution: An engine takes in 5 moles of air at 20 o C and 1 atm, and…

An engine takes in 5 moles of air at 20oC and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is________.

Step-by-Step Verified Solution

For diatomic ideal gas :

f = 5

γ\gamma = 75{7 \over 5}

Ti = T = 273 + 20 = 293 K

Vi = V

Vf = V10{V \over {10}}

For adiabatic process TV γ\gamma - 1 = constant

T1V1γ−1=T2V2γ−1{T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}

⇒\Rightarrow (293)V75−1=T2(V10)75−1\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}

⇒\Rightarrow T2=293×(10)25{T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}

Δ\Delta U = nfR(T2−T1)2{{nfR\left( {{T_2} - {T_1}} \right)} \over 2}

= 5×5×253×(293.1025−293)2{{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}

= 625×293×(1025−1)6{{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}

= 46.14 ×\times 103 J

≃\simeq 46 kJ

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