Question

If 1.0 kcal of heat is added to 1.2 L of $O_2$ in a cylinder of constant pressure of 1 atm, the volume increases to 1.5 L. Calculate $\Delta E$ and $\Delta H$ of the process:

• A. $\Delta E=0.993kcal,\Delta H=1kcal$
• B. $\Delta E=-0.993kcal,\Delta H=1kcal$
• C. $\Delta E=0.993kcal,\Delta H=-1kcal$
• D. None of these

Answer 1

Answer: (A)

$\Delta E=0.993kcal,\Delta H=1kcal$

The enthalpy change for the process is equal to the amount of heat added. $\Delta H=1kcal$

The enthalpy change is related tot he change in the internal energy by the relation $\Delta H=\Delta E+P\Delta V$

Substitute values in the above expression.

$1\times {10}^3\times 4.18=\Delta E+1.013\times {10}^5\times 3\times {10}^{-3}$

$\Delta E=(4180-303.9)Joule=\left(\frac{4149.61}{4.18}\right)cal$

Hence, the change in the internal energy is $\Delta E=0.993kcal$

Note: calories are converted to J by using the expression 1 cal $=4.18J$