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Question: An ideal gas expands from a volume of 6\(d{m^3}\) to 16\(d{m^3}\) against the constant external pres...

An ideal gas expands from a volume of 6dm3d{m^3} to 16dm3d{m^3} against the constant external pressure of 2.026×105Nm−22.026 \times {10^5}N{m^{ - 2}} . Find the enthalpy change if Δu\Delta u is 418 J.

Explanation

Solution

We can find the change in enthalpy of the given reaction by following the formula.
ΔH=Δu+W\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 6dm3d{m^3} to 16dm3d{m^3} .
We know that 1dm3d{m^3} = 10−3m3{10^{ - 3}}{m^3}
So, we can write that 6 dm3d{m^3} = 6×10−3m36 \times {10^{ - 3}}{m^3} and 16dm3=16×10−3m3d{m^3} = 16 \times {10^{ - 3}}{m^3}
Now, we can find the change in volume, ΔV\Delta V for this change.
So, ΔV=V2−V1\Delta V = {V_2} - {V_1}
ΔV=16×10−3−6×10−3=10×10−3=10−2m3\Delta V = 16 \times {10^{ - 3}} - 6 \times {10^{ - 3}} = 10 \times {10^{ - 3}} = {10^{ - 2}}{m^3}
Thus, we obtained that ΔV=10−2m3\Delta V = {10^{ - 2}}{m^3}
Now, we can find the work done by using a given formula.
W=PΔVW = P\Delta V
Here, W is work done, P is the external pressure of the system and ΔV\Delta V is the change in volume of the system.
So, we can put the available values into above equation to obtain
W=2.026×105×10−2W = 2.026 \times {10^5} \times {10^{ - 2}}
Thus, work done W = 2.026×103J2.026 \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.
ΔH=Δu+W (1)\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 = 2.026×103J2.026 \times {10^3}J .
We are given that Δu\Delta u =418 J. So, as we put all the available values into equation (1), we get
ΔH=418+2.026×103=418+2026=2444J\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 m3{m^3} as the value of internal energy Δu\Delta u is also given in the SI unit. Remember that 1dm3d{m^3} = 10−3m3{10^{ - 3}}{m^3}. Please note that the SI unit of both enthalpy and work is Joules.