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Question: An ideal gas expands from a volume of \(6{\text{ d}}{{\text{m}}^3}\) to \(16{\text{ d}}{{\text{m}}^3...

An ideal gas expands from a volume of 6 dm36{\text{ d}}{{\text{m}}^3} to 16 dm316{\text{ d}}{{\text{m}}^3} against the constant external pressure of2.026×105 Nm−22.026 \times {10^5}{\text{ N}}{{\text{m}}^{ - 2}}. Find the enthalpy change if ΔU\Delta U is418 J418{\text{ J}}.

Explanation

Solution

First calculate the work done by the ideal gas as it expands using the formula,
Work done =−PΔV - P\Delta V where P is pressure and ΔV\Delta V is change in volume. After finding the work done, calculate the change in enthalpyUsing the formula,
⇒ΔH = ΔU + PΔV\Rightarrow \Delta H{\text{ = }}\Delta U{\text{ + }}P\Delta V
Where ΔH\Delta H is enthalpy change, ΔU\Delta U is change in internal energy, P is pressure and ΔV\Delta V is change in volume.

Step-by-Step Solution-
Given the volume of ideal gas changes from 6 dm36{\text{ d}}{{\text{m}}^3} to 16 dm316{\text{ d}}{{\text{m}}^3}
And we know that, dm=10−1m{\text{dm}} = {10^{ - 1}}m
Then V1=6×(0.1m)3=6×10−3m3{V_1} = 6 \times {\left( {0.1m} \right)^3} = 6 \times {10^{ - 3}}{m^3}
And V2=16×(0.1m)3=16×10−3m3{V_2} = 16 \times {\left( {0.1m} \right)^3} = 16 \times {10^{ - 3}}{m^3}
Given the constant external pressure P=2.026×105 Nm−22.026 \times {10^5}{\text{ N}}{{\text{m}}^{ - 2}}
And ΔU\Delta U=418 J418{\text{ J}}
Now we know that the formula of work done =−PΔV - P\Delta V where P is pressure and ΔV\Delta V is change in volume
On putting the given values in the formula we get,
Work done=−2.026×105×(V2−V1) - 2.026 \times {10^5} \times \left( {{V_2} - {V_1}} \right)
On putting the values of initial and final Volume we get,
Work done=−2.026×105×(6×10−3−16×10−3) - 2.026 \times {10^5} \times \left( {6 \times {{10}^{ - 3}} - 16 \times {{10}^{ - 3}}} \right)
On solving the values inside the bracket we het,
Work done=−2.026×105×(−10×10−3) - 2.026 \times {10^5} \times \left( { - 10 \times {{10}^{ - 3}}} \right)
On multiplication we have,
Work done=2.026×10−3+1+52.026 \times {10^{ - 3 + 1 + 5}} {as we know thataman=am+n{a^m}{a^n} = {a^{m + n}} }
On solving we get,
Work done=2.026×103 J2.026 \times {10^3}{\text{ J}}
Now we know the work done so we can calculate the enthalpy change using the formula-
⇒ΔH = ΔU + PΔV\Rightarrow \Delta H{\text{ = }}\Delta U{\text{ + }}P\Delta V
Where ΔH\Delta H is enthalpy change , ΔU\Delta U is change in internal energy and P is pressure and ΔV\Delta V is change in volume .We have already calculated the last term.
Now putting the given values in the formula we get,
⇒ΔH = 418 + 2.026×103 \Rightarrow \Delta H{\text{ = }}418{\text{ + }}2.026 \times {10^3}{\text{ }}
On solving we get,
⇒ΔH = 418 + 2026 \Rightarrow \Delta H{\text{ = }}418{\text{ + }}2026{\text{ }}
On addition we have,
⇒ΔH = 2444 J\Rightarrow \Delta H{\text{ = }}2444{\text{ J}}

Answer-Hence enthalpy change is 2444 J2444{\text{ J}}

Note: The reason of the work done sign to be negative can be explained by the Important sign convention about the work done-
When work is done on the system by providing external pressure then the work done is positive.
When the work is done by the system by volume expansion then the work done is negative.