A gas is compressed at a constant pressure of 0.800 atm from 6.00 L to 1.00 L. In the process, 380 J of energy leaves the gas by heat.
(a) What is the work done on the gas?
(b) What is the change in its internal energy?
W= - p•ΔV = -0.8•(-5)•10⁻³•1.013•10⁵ = =405.2 J
ΔU =Q+W=-380+405.2=252 J
Student
1.013*10 5
(a) Ah, compressing gas, the ultimate workout for molecules! To find the work done on the gas, we can use the formula:
Work = -P(V2 - V1)
Where P is the constant pressure, and V2 and V1 are the initial and final volumes, respectively.
Plugging in the values, we get:
Work = -0.800 atm * (1.00 L - 6.00 L)
Work = -0.800 atm * (-5.00 L)
Work = 4.00 atm*L
But wait, we're not done clowning around yet! Since you didn't specify the units you want, we can convert this to joules by using the handy dandy conversion factor:
1 atm*L = 101.325 J
So, the work done on the gas is approximately 4.00 atm*L * 101.325 J/atm*L = 405.3 J.
(b) Now, let's move on to the change in internal energy. The change in internal energy (delta U) is given by the sum of the heat transferred (Q) and the work done on the gas (W):
ΔU = Q + W
Here, the heat transferred is stated as -380 J because it's leaving the gas. So, the change in internal energy would be:
ΔU = -380 J + 405.3 J
ΔU = 25.3 J
Oh, the poor gas must feel quite drained after all that work and heat loss!
To find the work done on the gas, we can use the equation:
Work = Pressure * Change in Volume
(a) The pressure of the gas is given as 0.800 atm, and the change in volume is from 6.00 L to 1.00 L. So the change in volume is:
Change in Volume = Final Volume - Initial Volume
= 1.00 L - 6.00 L
= -5.00 L
Note that the change in volume is negative because the gas is being compressed. Now we can calculate the work done:
Work = Pressure * Change in Volume
= 0.800 atm * (-5.00 L)
To complete the calculation, we need to convert atm (atmospheres) to a more appropriate unit for work, such as joules (J). The conversion factor is:
1 atm = 101.325 J
So:
Work = 0.800 atm * (-5.00 L) * 101.325 J/atm
Now we can compute the work:
Work = -404.7 J
Therefore, the work done on the gas is -404.7 J (negative because work is done on the system when the volume is decreasing).
(b) The change in internal energy, ΔU, is related to the heat transfer, q, and the work done, w, by the equation:
ΔU = q - w
Given that 380 J of energy leaves the gas by heat, and the work done on the gas is -404.7 J, we can calculate the change in internal energy:
ΔU = 380 J - (-404.7 J)
Simplifying this expression gives:
ΔU = 380 J + 404.7 J
ΔU = 784.7 J
Therefore, the change in internal energy is 784.7 J.