Question

Calculate the change in internal energy if $\Delta H = - 92.2 kJ$, $P = 40 \ atm$ and $\Delta V= - 1 \ L$
(a) - 42 kJ
(b) - 88 kJ
(c) + 88 kJ
(d) + 42 kJ

Answer 1

Hint: Internal energy is the energy associated with the random or disordered motions of the molecules. The change in internal energy is calculated by the difference of final internal energy and initial internal energy. It can be positive or negative.

Complete step by step answer:
Given in the question:
Enthalpy of the system i.e. $\Delta H=-92.2kJ$
Pressure = 40 atm
Change in volume i.e. $\Delta V=-1 \ L$
Not the formulae for calculation of enthalpy of a system
$\Delta H=\Delta E+P\Delta V$
Where $\Delta H$ is the enthalpy of the system, $P$ is the pressure, $\Delta V$ is the change in volume of the system and $\Delta E$ is the change in internal energy.
We have to calculate the change in internal energy,
$\Delta E=\Delta H+P\Delta V$
Put all the given values in the above equation, we get,
$\Delta E$ = \[-92.2\text{ }\text{ }40\left( -1 \right)\]
$\Delta E$= \[-92.2kJ\text{ + }40atmL\]
Now the volume and pressure is given in litre and atm respectively so we have to change it into joule.
We multiply and divide \[\text{ }40atmL\] with universal gas constant to convert it into joule and multiply \[-92.2kJ\] into joule.
1kJ = 1000 J
 $\Delta E$= \[-92.2kJ(1000) \text{ + }40atmL\dfrac{(8.314J{{K}^{-1}}mo{{l}^{-1}}) }{(0.0821atmL{{K}^{-1}}mo{{l}^{-1}})}\]
$\Delta E$ = -88149.34 J
$\Delta E$= -88.149 kJ
$\Delta E$= -88 kJ (-88.149kJ is approximately equal to -88kJ)

Hence, the correct answer is option (B).
The change in internal energy if $\Delta H=-92.2 \ kJ, \ P = 40 \ atm$ and $\Delta V$ = -1L is $-88 \ kJ$.

Note: If the value of $\Delta H$ is positive, then the reaction is exothermic which means that the heat is absorbed by the system and when the value of $\Delta H$ is negative, then the reaction is endothermic which means that the heat is released by the system.