The figure shows a cube of air at pressure 1.0 atm above a hot surface on Earth. Contact with the hot surface causes the volume of the air to expand from 2.00 × 105 L to 2.20 × 105 L by the transfer of 8.2 × 106 J of heat. Assume the systems are isolated from the outside world but not from each other.
b. Determine the change in energy for the air. Show your calculations. (Use the conversion 1 L· atm = 101.3 J.)
Delta E = q + w
W = -p * delta V
W = -1 atm * 2.0 * 10^4 = -2.0 * 10^4 L*atm * 101.3 J= 2026000 J
Delta E = 2026000 J *8.2 × 106 J = 1.66 * 10^13 J.
I was told that I made some calculation errors please help me find them.
The process is correct but since W is negative it is supposed to be -2026000 J.
And for part be the equation is Delta E = q+w but as w is negative it would be (-). So the proper equation would be 8.2x10^6 - 2026000J.
Well, it seems like there might be a few calculation errors in your work. Let's go through it step by step to find them.
First, let's calculate the work done on the air. You correctly used the equation W = -P * ΔV, but there's a small mistake in the units. The pressure is given as 1.0 atm, so the correct calculation would be:
W = -1.0 atm * (2.2 × 10^5 L - 2.0 × 10^5 L)
Now let's convert the units. Since 1 L· atm = 101.3 J, the work done can be calculated as follows:
W = -1.0 atm * (2.0 × 10^5 L - 2.2 × 10^5 L) * 101.3 J
Next, let's find the correct value of W:
W = -1.0 atm * (-2.0 × 10^4 L) * 101.3 J = 2.026 × 10^6 J
Now, let's calculate the change in energy for the air using the correct value of W:
ΔE = q + W = 8.2 × 10^6 J + 2.026 × 10^6 J = 1.0226 × 10^7 J
So, the correct change in energy for the air is approximately 1.0226 × 10^7 J.
Hope that helps!
In order to determine the change in energy for the air, you need to calculate the work done on the air and then add it to the heat absorbed by the air.
Let's start by calculating the work done on the air. The work done can be calculated using the formula:
W = -P * ΔV
where:
W represents the work done on the air
P represents the pressure on the air
ΔV represents the change in volume of the air
In this case, the pressure on the air is given as 1.0 atm, and the change in volume is given as 2.20 × 10^5 L - 2.00 × 10^5 L = 2.0 × 10^4 L.
Now let's calculate the work done on the air:
W = -1.0 atm * 2.0 × 10^4 L * 101.3 J / (1 L · atm) = -2.026 × 10^6 J
Next, we can calculate the change in energy (ΔE) for the air:
ΔE = q + W
where:
ΔE represents the change in energy
q represents the heat absorbed by the air
W represents the work done on the air
In this case, the heat absorbed by the air is given as 8.2 × 10^6 J.
Now let's calculate the change in energy for the air:
ΔE = 8.2 × 10^6 J + (-2.026 × 10^6 J) = 6.174 × 10^6 J
Therefore, the correct calculation for the change in energy for the air is ΔE = 6.174 × 10^6 J, not 1.66 × 10^13 J as previously calculated.
You have tried it with correct approach. Rest was perfect but the multiplication sign.
Determine the change in energy for the air. Show your calculations. (Use the conversion 1 L· atm = 101.3 J.)
Delta E = q + w
W = -p * delta V
W = -1 atm * 2.0 * 10^4 = -2.0 * 10^4 L*atm * 101.3 J= 2026000 J
Delta E = 2026000 J *8.2 + 106 J = -19894 J
Or -1.98 x 10^4 J