Question

Three mole of $\mathrm{Ag}$ is heated from $300Kto1000K$. Calculate $\Delta H$ when $P=1atmand{C}_P=23+0.01T$.

• A. $62\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$
• B. $45\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$
• C. $54\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$
• D. $38\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Answer 1

The correct option is A

$62\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Step 1: Given information

1. Mole of $\mathrm{Ag}$ is heated=$300Kto1000K$
2. $P=1atm$
3. $C_P=23+0.01T$

Step 2: Explanation

Knows, $\Delta H={\int }_{T_1}^{T_2}nCpdT$, Where, $C_p$= the amount of heat energy released or absorbed, $T$= Temprature and $∆H$= Total heat

Therefore,

$\Rightarrow \Delta H=n{\int }_{300}^{1000}\left(23+0.1T\right)dT$

$\Rightarrow \Delta H=3{\left[23T+\frac{0.01T^2}{2}\right]}_{300}^{1000}$

Substitute limits,

$$\begin{gathered} \Rightarrow \Delta H=3\left[\frac{2\times 23\times 700+9100}{2}\right] \\ \Rightarrow \Delta H=61960\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \\ \Rightarrow \Delta H=62kJ/mol \end{gathered}$$

Hence, the correct option is (A).