Pt. 1
Gas in a container is at a pressure of 2.3 atm
and a volume of 6.1 m3.
What is the work done on the gas if it
expands at constant pressure to five times its
initial volume?
Answer in units of J
My wrong answer: 56.12
Pt. 2:
What is the work done on the gas if it is com-
pressed at constant pressure to one-quarter of
its initial volume?
Answer in units of J
I'll help with part 1
W=P(delta)V or work=pressure X volume change
W=(2.3 atm )(5 x 6.1)
W=70.15 J
To calculate the work done on the gas, you can use the formula:
Work = Pressure * Change in Volume
Let's solve both parts of the question step by step.
Part 1:
Given:
Pressure of the gas (P) = 2.3 atm
Initial volume (V1) = 6.1 m^3
Final volume (V2) = 5 times V1
To find the change in volume (ΔV), subtract the initial volume from the final volume:
ΔV = V2 - V1 = (5 * V1) - V1 = 4 * V1
Now substitute the values into the formula:
Work = P * ΔV = 2.3 atm * 4 * 6.1 m^3
To ensure the final answer is in Joules (J), we need to convert atm to pascals (Pa) since 1 Pa = 1 N/m^2 and 1 J = 1 N * m:
1 atm = 101325 Pa.
Therefore:
Work = 2.3 * 101325 Pa * 4 * 6.1 m^3 = 5,605,207 J (rounded to the nearest whole number).
So, the correct answer for Part 1 is 5,605,207 J.
Part 2:
Given:
Pressure of the gas (P) = 2.3 atm
Initial volume (V1) = 6.1 m^3
Final volume (V2) = one-quarter of V1
To find the change in volume (ΔV), subtract the final volume from the initial volume:
ΔV = V2 - V1 = (1/4) * V1 - V1 = -(3/4) * V1
Now substitute the values into the formula:
Work = P * ΔV = 2.3 atm * (-3/4) * 6.1 m^3
Converting atm to pascals:
Work = 2.3 * 101325 Pa * (-3/4) * 6.1 m^3 = -1,292,492 J (rounded to the nearest whole number).
So, the correct answer for Part 2 is -1,292,492 J.
Note: The negative sign indicates that work is done on the gas during compression, as energy is transferred from the surroundings to the system.