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Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q46

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Heat and Thermodynamics
2023 · 31 Jan · Shift 2 · Q46

Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q46

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

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 :

A
$572 \mathrm{~J}$
B
$441 \mathrm{~J}$
C
$525 \mathrm{~J}$
D
$735 \mathrm{~J}$

View written solutionFree

Correct answer: C

Identify the degrees of freedom

For a diatomic gas in this condition:

translational degrees of freedom $=3$
rotational degrees of freedom $=2$ (about two axes perpendicular to the bond)
vibrational degrees of freedom are neglected

So total active degrees of freedom: $f=3+2=5$

Hence, $C_V=\frac{f}{2}R=\frac{5}{2}R$

and $C_P=C_V+R=\frac{7}{2}R$

Use the relation for heat supplied at constant pressure

At constant pressure, $Q=nC_P\Delta T$

The increase in internal energy is $\Delta U=nC_V\Delta T$

Therefore, $\Delta U=\frac{C_V}{C_P}Q$

Substitute the values:

=\frac{5}{7}\times 735$$ 3. **Calculate** $$\Delta U=5\times 105=525\,\text{J}$$ 4. **Check options** - A: $572\,\text{J}$ ❌ - B: $441\,\text{J}$ ❌ - C: $525\,\text{J}$ ✅ - D: $735\,\text{J}$ ❌ So the correct answer is **Option C**.

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