Question

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 6.40 to 3.20 L . When the external pressure is increased to 2.50 atm , the gas further compresses from 3.20 to 2.56 L .

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 6.40 to 2.56 L in one step.

If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

Answer 1

Answer: -162,120J

Step-by-step explanation:

Okay, so q stands for work, given by the formula q = -P∆V,

For the first round, we have 2atm (for P) and a volume change of 3.2 (6.4-3.2) for the first experiment.

Find the work: q = -2 • (3.2 - 6.4) = 6.4L/atm

For the second experiment, it is now 2.5atm, with the change in volume going from 3.2 to 2.56.

So, work is q = -2.5(2.5 - 3.20) = 1.6L/atm

Add the total work: 1.6 + 6.4 = 8L/atm

To find the Joules per L/atm, recall that a mol of gas in K is 8.31447J, while the gas constant where a liter of gas per mol in K is 0.08206L/atmK. Divide these two values, which leaves us with 101,325J

So, for the first round, multiply 8 x 101,325, which leaves us with 810,600J.

Second round: The pressure is 2.5atm, with the volume change from 6.40 to 2.56

Solve for work: q = -2.5(2.56-6.4) = 9.6L/atm

Multiply with the Joule unit found earlier: 9.6 x 101,325 = 972,720J.

To find the difference, subtract the value of the single-step process (second round) from the first round (multistep)

810600 - 972720 = -162,120J. Therefore, it takes more heat to do the single step than the multistep process.