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III. Long Answer Questions · Q5

Q.Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules.

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Step 1. Monatomic (f=3). U=32RTU=\tfrac32RT, so CV=dU/dT=32RC_V=dU/dT=\tfrac32R. By Meyer's relation, CP=CV+R=52RC_P=C_V+R=\tfrac52R. So γ=5/23/2=53≈1.67\gamma=\dfrac{5/2}{3/2}=\dfrac53\approx1.67.

Step 2. Diatomic, normal temperature (f=5). U=52RTU=\tfrac52RT, CV=52RC_V=\tfrac52R, CP=72RC_P=\tfrac72R, γ=7/25/2=75=1.40\gamma=\dfrac{7/2}{5/2}=\dfrac75=1.40.

Step 3. Diatomic, high temperature (f=7). U=72RTU=\tfrac72RT, CV=72RC_V=\tfrac72R, CP=92RC_P=\tfrac92R, γ=9/27/2=97≈1.28\gamma=\dfrac{9/2}{7/2}=\dfrac97\approx1.28.

Step 4. Triatomic, linear (f=7). Same as diatomic high-temperature case: CV=72RC_V=\tfrac72R, CP=92RC_P=\tfrac92R, γ=97≈1.28\gamma=\tfrac97\approx1.28.

Step 5. Triatomic, non-linear (f=6). U=3RTU=3RT, CV=dU/dT=3RC_V=dU/dT=3R, CP=3R+R=4RC_P=3R+R=4R, γ=43≈1.33\gamma=\dfrac43\approx1.33. …

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