JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 2)
Heat energy of $735 \mathrm{~J}$ is given to a diatomic gas allowing the gas to expand at constant pressure.
Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal
energy of the gas will be :
$572 \mathrm{~J}$
$441 \mathrm{~J}$
$525 \mathrm{~J}$
$735 \mathrm{~J}$
Explanation
$\Delta Q=n C_{P} \Delta T= $
$$n\left( {{f \over 2} + 1} \right)R\Delta T$$
= $$n\left( {{5 \over 2} + 1} \right)R\Delta T$$
= $$n\left( {{7 \over 2}} \right)R\Delta T$$
Given, $\Delta Q$ = 735
$$ \Rightarrow $$ $$n\left( {{7 \over 2}} \right)R\Delta T$$ = 735
$$ \Rightarrow $$ $$nR\Delta T = 735 \times {2 \over 7}$$
Also, $$\Delta $$U = $${f \over 2}nR\Delta T$$
= $${5 \over 2}nR\Delta T$$
= $${5 \over 2} \times 735 \times {2 \over 7}$$
= 525 J
$$n\left( {{f \over 2} + 1} \right)R\Delta T$$
= $$n\left( {{5 \over 2} + 1} \right)R\Delta T$$
= $$n\left( {{7 \over 2}} \right)R\Delta T$$
Given, $\Delta Q$ = 735
$$ \Rightarrow $$ $$n\left( {{7 \over 2}} \right)R\Delta T$$ = 735
$$ \Rightarrow $$ $$nR\Delta T = 735 \times {2 \over 7}$$
Also, $$\Delta $$U = $${f \over 2}nR\Delta T$$
= $${5 \over 2}nR\Delta T$$
= $${5 \over 2} \times 735 \times {2 \over 7}$$
= 525 J
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