For silver,C_P , (J , K^{-1} , mol^{-1}) = 23 | vedclass

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(B) The change in enthalpy $(\Delta H)$ for a process at constant pressure is given by the integral of $n \, C_P \, dT$. Given $n = 3 \, mol$,$C_P = 2

Vedclass · Accepted Answer

$62$

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(B) The change in enthalpy $(\Delta H)$ for a process at constant pressure is given by the integral of $n \, C_P \, dT$. Given $n = 3 \, mol$,$C_P = 23 + 0.01 \, T$,$T_1 = 300 \, K$,and $T_2 = 1000 \, K$. $\Delta H = n \int_{T_1}^{T_2} C_P \, dT = 3 \int_{300}^{1000} (23 + 0.01 \, T) \, dT$. $\Delta H = 3 [23 \, T + \frac{0.01 \, T^2}{2}]_{300}^{1000}$. $\Delta H = 3 [23(1000 - 300) + 0.005(1000^2 - 300^2)]$. $\Delta H = 3 [23(700) + 0.005(1000000 - 90000)]$. $\Delta H = 3 [16100 + 0.005(910000)]$. $\Delta H = 3 [16100 + 4550] = 3 [20650] = 61950 \, J$. Converting to $kJ$,$\Delta H = 61.95 \, kJ \approx 62 \, kJ$.

For silver,$C_P \, (J \, K^{-1} \, mol^{-1}) = 23 + 0.01 \, T$. If the temperature $(T)$ of $3 \, moles$ of silver is raised from $300 \, K$ to $1000 \, K$ at $1 \, atm$ pressure,the value of $\Delta H$ will be close to $kJ$.

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