Two moles of air, when heated through 10 K expand by an amount of 1.66×10−3m3 . Under a constant pressure of 105N/m2 . If CV=20.81 J/mol/K , then CP is ____J/mol/K

ΔW=P×ΔV=105×1.66×10−3 J⇒ΔW=166 JΔU=nCVΔT=2×20.81×10 J⇒ΔU=416.2 JNow, ΔQ=ΔU+ΔW=(416.2+166)J=582.2JAlso, ΔQ=nCPΔT∴CP=ΔQnΔT=582.22×10J/mol−K=29.1 J/mol−K

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Two moles of air, when heated through 10 K expand by an amount of $1.66 \times 10^{- 3} m^{3}$ . Under a constant pressure of $10^{5} N / m^{2}$ . If $C_{V} = 20.81 J / m o l / K$ , then $C_{P}$ is $____J / m o l / K$

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answer is 29.

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Detailed Solution

$\begin{array}{l} ΔW = P \times ΔV = 10^{5} \times 1.66 \times 10^{- 3} J \\ \Rightarrow ΔW = 166 J \\ ΔU = {nC}_{V} ΔT = 2 \times 20.81 \times 10 J \\ \Rightarrow ΔU = 416.2 J \end{array}$