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
To find the work done on the gas during expansion or compression, you can use the formula:
Work = Pressure * Change in Volume
In Part 1 of the question, the gas is expanding at constant pressure to five times its initial volume.
Change in Volume = Final Volume - Initial Volume
= 5 * Initial Volume - Initial Volume
= 4 * Initial Volume
So, the work done on the gas during this expansion is given by:
Work = Pressure * Change in Volume
= 2.3 atm * 4 * 6.1 m^3
To solve this equation, you need to convert the pressure from atm (atmospheres) to SI units of Pascal (Pa). 1 atm is equal to 101325 Pa.
So, the work can be calculated as:
Work = (2.3 atm * 101325 Pa/atm) * 4 * 6.1 m^3
Now, let's calculate the answer:
Work = 2.3 * 101325 * 4 * 6.1
And the answer is in units of Joules (J).
Now, let's move on to Part 2 of the question.
In Part 2, the gas is compressed at constant pressure to one-quarter of its initial volume.
Change in Volume = Final Volume - Initial Volume
= 1/4 * Initial Volume - Initial Volume
= -3/4 * Initial Volume
Since the volume decreases with compression, the value will be negative.
The work done on the gas during this compression is:
Work = Pressure * Change in Volume
= 2.3 atm * (-3/4) * Initial Volume
Now, convert the pressure to the SI unit of Pascal (Pa) as explained earlier.
Work = (2.3 atm * 101325 Pa/atm) * (-3/4) * Initial Volume
Calculate the answer by substituting the values:
Work = 2.3 * 101325 * (-3/4) * Initial Volume
And the answer is in units of Joules (J).