JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 23)
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________.
Answer
46
Explanation
For diatomic ideal gas :
f = 5
$$\gamma $$ = $${7 \over 5}$$
Ti = T = 273 + 20 = 293 K
Vi = V
Vf = $${V \over {10}}$$
For adiabatic process TV$$\gamma $$ - 1 = constant
$${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$
$$ \Rightarrow $$ $$\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$$
$$ \Rightarrow $$ $${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$$
$$\Delta $$U = $${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$$
= $${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$$
= $${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$$
= 46.14 $$ \times $$ 103 J
$$ \simeq $$ 46 kJ
f = 5
$$\gamma $$ = $${7 \over 5}$$
Ti = T = 273 + 20 = 293 K
Vi = V
Vf = $${V \over {10}}$$
For adiabatic process TV$$\gamma $$ - 1 = constant
$${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$
$$ \Rightarrow $$ $$\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$$
$$ \Rightarrow $$ $${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$$
$$\Delta $$U = $${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$$
= $${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$$
= $${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$$
= 46.14 $$ \times $$ 103 J
$$ \simeq $$ 46 kJ
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