Thermodynamics

JEE Main level, numerical answer, medium

22 of 190

Question 22 of 190

10 mole of oxygen is heated at constant volume from 30∘C30^{\circ} \mathrm{C} to 40∘C40^{\circ} \mathrm{C}. The change in the internal energy of the gas is ____\_\_\_\_ cal. (The molecular specific heat of oxygen at constant pressure, CP=7cal/mol.∘CC_{P}=7 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C} and R=2cal/mol.∘C\mathrm{R}=2 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C}.)

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Answer: 500

Given: n=10n = 10 mol,
ΔT=40−30=10°C\Delta T = 40 - 30 = 10°C,
CP=7C_P = 7 cal/mol·°C,
R=2R = 2 cal/mol·°C
For ideal gas: CV=CP−R=7−2=5C_V = C_P - R = 7 - 2 = 5 cal/mol·°C
Change in internal energy: ΔU=nCVΔT=10×5×10=500\Delta U = nC_V\Delta T = 10 \times 5 \times 10 = 500 cal

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