A gaseous reaction occurs at a constant pressure of 50.0 atm and releases 55.6 kJ of heat. Before the reaction, the volume of the system was 7.60 L. After the reaction, the volume of the system was 2.00 L.
Calculate the total internal energy change, Delta U, in kilojoules.
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To find the total internal energy change, ΔU, we can use the equation:
ΔU = q + w
Where:
- ΔU is the change in internal energy
- q is the heat transferred
- w is the work done
In this case, the reaction releases heat, so q will be negative. The work done can be calculated using the formula:
w = -PΔV
Where:
- w is the work done
- P is the pressure
- ΔV is the change in volume
Given:
- q = -55.6 kJ
- P = 50.0 atm
- ΔV = 7.60 L - 2.00 L = 5.60 L
First, we need to ensure that the units are consistent. We can convert L to atm by using 1 L = 0.101325 atm:
ΔV = 5.60 L x 0.101325 atm/L = 0.56616 atm
Now we can calculate the work done:
w = -PΔV
w = -(50.0 atm)(0.56616 atm) = -28.308 kJ
Finally, we can calculate the total internal energy change:
ΔU = q + w
ΔU = -55.6 kJ - 28.308 kJ
ΔU = -83.908 kJ
Therefore, the total internal energy change, ΔU, is -83.908 kJ.
To calculate the total internal energy change (ΔU), you can use the equation:
ΔU = q + w
where ΔU is the change in internal energy, q is the heat transferred to or from the system, and w is the work done by or on the system.
In this case, you are given q, the heat released by the reaction, which is -55.6 kJ (the negative sign indicates heat released). Since it is released by the system, it will be negative.
Next, you need to calculate the work done by the system. The work can be calculated using the formula:
w = -PΔV
where P is the pressure and ΔV is the change in volume.
Given:
Pressure (P) = 50.0 atm
Initial volume (V₁) = 7.60 L
Final volume (V₂) = 2.00 L
ΔV can be calculated by subtracting the final volume from the initial volume:
ΔV = V₂ - V₁
ΔV = 2.00 L - 7.60 L
ΔV = -5.60 L
(Note: The negative sign indicates a decrease in volume.)
Now, substitute the values into the formula for work:
w = -PΔV
w = -(50.0 atm)(-5.60 L)
w = 280 atm·L
Finally, substitute both the heat (q) and work (w) values into the equation for ΔU:
ΔU = q + w
ΔU = -55.6 kJ + 280 atm·L
Note that we have the units of energy (kJ) and work (atm·L), so we need to convert the units of work to match the energy units. The conversion can be done using the following relationship:
1 atm·L = 101.325 J
Therefore, multiply the work value by the conversion factor:
w = 280 atm·L × 101.325 J/1 atm·L
w ≈ 28357.8 J
Now, convert the work value from joules to kilojoules:
w ≈ 28.36 kJ
Finally, substitute the converted values into the equation for ΔU:
ΔU = -55.6 kJ + 28.36 kJ
ΔU ≈ -27.24 kJ
Therefore, the total internal energy change (ΔU) in kilojoules is approximately -27.24 kJ.