Thermodynamics

JEE Main level, numerical answer, medium

65 of 190

Question 65 of 190

When 300 J of heat given to an ideal gas with Cp=72RC_{p}=\frac{7}{2} R its temperature raises from 20∘C20^{\circ} \mathrm{C} to 50∘C50^{\circ} \mathrm{C} keeping its volume constant. If n is the number of moles of the gas, then what is the value of 100n100n? (R=8.314 J/mol.K)(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol}. \mathrm{K})

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Answer: 48

Given: CP=72RC_P = \frac{7}{2}R, so CV=CP−R=52RC_V = C_P - R = \frac{5}{2}R
ΔQ=nCVΔT\Delta Q = nC_V \Delta T (constant volume)
300=n×52×8.314×30300 = n \times \frac{5}{2} \times 8.314 \times 30
n=0.48n = 0.48 mol
100n=48100n = 48

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