Question

One mole of non – ideal gas undergoes a change of state (1.0 atm, 3.0 L, 200 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy \[\Delta U = 40L - atm\]. The change in enthalpy of the process in L – atm is:
A.43
B.57
C.42
D.None of these

Answer 1

Hint: In any given reaction, the heat evolved from the process is equivalent to the change in the enthalpy of the system. To solve this question, we need to write down all the given data. Then we must substitute the given data in the equation for the law of conservation of energy to get the final answer.
Formula used: \[H{\text{ }} = {\text{ }}U{\text{ }} + {\text{ }}PV\].

Complete step by step answer:
Enthalpy (H) is mathematically equal to the sum of the internal energy (U) of the system and the product of pressure (P) and volume (V) of the system. This can be represented as:
 \[H{\text{ }} = {\text{ }}U{\text{ }} + {\text{ }}PV\]
This is also known as the law of conservation of energy. Hence, the change in the enthalpy of the system involves the change in the values of all the components in the above equation. Hence, the change in enthalpy can be represented as:
 \[\Delta H = \Delta U + \Delta (PV)\]
 \[\Delta H = \Delta U + [{(\Delta PV)_2} - {(\Delta PV)_1}]\]
 \[\Delta H = 40{\text{ }} + \left[ {\left( {4.0} \right){\text{ }}\left( {5.0} \right) - \left( {1.0} \right){\text{ }}\left( {3.0} \right)} \right]\]
 \[\;\Delta H = 40{\text{ + }}\left[ {17} \right]\]
 \[\Delta H = 57{\text{ }}L - atm\]

Hence, Option B is the correct option.

Additional Information:
The enthalpy of a reaction can be categorised into various types – heat of formation, heat of combustion, heat of neutralization and heat of solution. These different enthalpies depend on the objective of the reaction.

Note:
Enthalpy (H) is a state function which is dependent on the values of temperature, pressure, and internal energy. Enthalpy is usually expressed as a change enthalpy because absolute enthalpy of any system cannot be physically calculated.