First Law Applied To Constant Pressure Heating

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When 2 moles of a monatomic ideal gas are heated at constant pressure causing a temperature rise of 50 K, so how much heat must be supplied? Take R = 8.3 J per mol kelvin.

RankGuru · Accepted Answer

Based on NCERT Class 11, Chapter 12 (Thermodynamics), the heat supplied at constant pressure is $Q = n C_P \Delta T$, where for a monatomic ideal gas $C_V = \dfrac{3R}{2}$ and by Mayer's relation $C_P = C_V + R = \dfrac{3R}{2} + R = \dfrac{5R}{2}$. Substituting $n = 2$, $\Delta T = 50$ K and $R = 8.3$: $C_P = \dfrac{5 	imes 8.3}{2} = 20.75$ J/mol·K, so $Q = 2 	imes 20.75 	imes 50 = 2075$ J. Part of this heat raises internal energy and part does expansion work. The option 1245 J is wrong because it uses $C_V = 3R/2$, computing only the internal energy change and ignoring the expansion work at constant pressure. The option 830 J is wrong because it uses $R$ alone times the relevant factors, omitting the specific heat structure. The option 4150 J is wrong because it doubles the correct value, likely using $5R$ instead of $5R/2$. A plausibility check: at constant pressure $Q$ must exceed the constant-volume heat $nC_V\Delta T = 1245$ J because extra energy goes into work, and indeed $2075 > 1245$, confirming the answer is consistent.

First Law Applied To Constant Pressure Heating

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