Question

An ideal gas expanded irreversibly from 5L to 10L against a constant external pressure of 1 bar. The value of heat involved (q) in this isenthalpic process is:
A. 0
B. +500 J
C. +5 J
D. -500 J

Answer 1

Hint: For this problem, we have to use the formula of work which is equal to the product of the pressure and change in the volume. After which we will apply the formula of heat for the isenthalpic process which is equal to the negative of work.

Complete step by step solution:
- in the given question, we have to calculate the heat that will be involved in the isenthalpic process by using the given data.
- It is given that the initial volume and final volume is equal to 5 L and 10 L respectively and the pressure is 1 bar which is also equal to the 0.9869 atm so the work will be:
$\text{W = -P}\Delta \text{V}$
- Here, the volume change is equal to the initial volume that is subtracted from the final volume.
$\text{W = -0}\text{.9869(10 - 5) = -4}\text{.9345 atmL}$
Or $\text{W = -4}\text{.9345 }\times \text{ 101}\text{.33 = -500J}$
- Now, as we know that the given process is an isenthalpic process which means that the change in the enthalpy is equal or constant.
- As we know that the first law of thermodynamics explains the conservation of energy that is the energy can neither be created nor destroyed and only transformed into another form.
- So, the first law of thermodynamics that is expressed by $\Delta \text{U = Q - W}$ will become:
$\text{-Q = W}$ … (1)
- Now, by putting the value of work in equation (1) we will get:
$Q =$ $+500 J$

Therefore, option B is the correct answer.

Note: The enthalpy is responsible for determining whether the system will absorb or release the heat while doing the work. So, if the value of enthalpy will be constant as in isenthalpic process it means that there is no exchange of heat in the system.