Question

An ideal gas expands from a volume of 6$d{m^3}$ to 16$d{m^3}$ against the constant external pressure of $2.026 \times {10^5}N{m^{ - 2}}$ . Find the enthalpy change if $\Delta u$ is 418 J.

Answer 1

Hint: We can find the change in enthalpy of the given reaction by following the formula.
\[\Delta H = \Delta u + W\]


Complete Step-by-Step Solution:
We will calculate the change in enthalpy by using the relation between enthalpy and internal energy.
- We are given that the change in volume is from 6$d{m^3}$ to 16$d{m^3}$ .
We know that 1$d{m^3}$ = ${10^{ - 3}}{m^3}$
So, we can write that 6 $d{m^3}$ = $6 \times {10^{ - 3}}{m^3}$ and 16$d{m^3} = 16 \times {10^{ - 3}}{m^3}$
Now, we can find the change in volume, $\Delta V$ for this change.
So, $\Delta V = {V_2} - {V_1}$
$\Delta V = 16 \times {10^{ - 3}} - 6 \times {10^{ - 3}} = 10 \times {10^{ - 3}} = {10^{ - 2}}{m^3}$
Thus, we obtained that $\Delta V = {10^{ - 2}}{m^3}$
Now, we can find the work done by using a given formula.
\[W = P\Delta V\]
Here, W is work done, P is the external pressure of the system and $\Delta V$ is the change in volume of the system.
So, we can put the available values into above equation to obtain
\[W = 2.026 \times {10^5} \times {10^{ - 2}}\]
Thus, work done W = $2.026 \times {10^3}J$
Now, we will use the thermodynamic relation between enthalpy, internal energy and work done. The equation can be given as follows.
\[\Delta H = \Delta u + W{\text{ }} {\text{(1)}}\]
Now, we already found that work done is equal to the product of pressure and the change in volume. So, we already obtained that W = $2.026 \times {10^3}J$ .
We are given that $\Delta u$ =418 J. So, as we put all the available values into equation (1), we get
\[\Delta H = 418 + 2.026 \times {10^3} = 418 + 2026 = 2444J\]

Thus, we obtained that the enthalpy change for the given reaction is 2444 J.


Note: Do not forget to convert the unit of volume into its SI unit ${m^3}$ as the value of internal energy $\Delta u$ is also given in the SI unit. Remember that 1$d{m^3}$ = ${10^{ - 3}}{m^3}$. Please note that the SI unit of both enthalpy and work is Joules.