Question

One mole of a non-ideal gas undergoes a change state $\left( {2 \;atm,3\;L,95\;K} \right)$ to $\left( {4 \;atm,5\;L,245\;K} \right)$ with a change of internal energy, $\Delta U = 30L\,atm$. The change in enthalpy $\Delta {\rm H}$ of the pressure in $L\;atm$:
A) 40
B) 42.3
C) 44.0
D) Not defined pressure is not constant

Answer 1

Hint: Use that expression of enthalpy change where all three factors pressure, volume, and temperature changes.
We know that every substance is associated with a definite amount of energy which depends upon its chemical nature as well as upon temperature, pressure and volume. This energy is known as internal energy. Internal energy of a substance is a definite quantity and it is the function of temperature only.

Complete step by step answer:
Suppose a system is subjected to change of pressure and volume. Let the initial state be represented by A and the final state is represented by B then the change in internal energy of both states will be;
$\Delta U = {U_B} - {U_A}$
The exact magnitude of this energy is not known because the chemical nature includes different factors like translational, rotational and vibrational movements of the movement of molecules, the manner in which the molecules are put together, the nature of the individual atoms, the arrangement and number of electrons, the energy possessed by the nucleus etc.
Given that-
$\begin{array}{l}
{P_1} = 2\,atm\\
{V_1} = 3L\\
{P_2} = 4\,atm\\
{V_2} = 5L\\
\Delta U = 30Latm
\end{array}$
Using the formula-
$\Delta H = \Delta U + \Delta {PV}$
$= 30 + {{P_2}{V_2} - {P_1}{V_1}}$
$= 30 + {20 - 6}$
$= 30 + 14$
$= 44\,L\;atm$
Hence, the change in enthalpy is 44 L atm.

Hence, the option C is correct.

Note: By conventions of IUPAC in chemical thermodynamics. The q is positive, when heat is transferred from the surroundings to the system and the internal energy of the system increases and q is negative when heat is transferred from system to the surroundings resulting in decrease of the internal energy of the system.