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 4.90 to 2.45L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.45 to 1.96L .

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 4.90 to 1.96L 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?

Express your answer with the appropriate units.

Answer 1

1) Calculate w for the one step process.

Take the change in volume 1.96L - 4.90 L = -2.94L.

(-2.94 L / 1000) (2.5atm*101325pascals) = 744.738 L atm K pa

744.738 / 101.3 = 7.35 Liter-atmospheres

Thus, W 1= 7.35 L-atm

2) Calculate w for the two-step process.

Basically, repeat the same process with different values:
2.45L - 4.90L = -2.45 L / 1000 = -0.00245L.

-0.00245L(101325pascals*2atm) = -496.49J

-496.49 J / 101.3 = -4.90 L atm

Since this is a two-step process, the total work is the sum of the work of the first step and the second step. Now, we will calculate the work done in the second step and sum them to get the total work done in this two-step process.

1.96L - 2.45L = -0.49 L / 1000 = -0.00049L

-0.00049L(2.5atm*101325pascals)= -124.123 J

-124.123 J / 101.3 = -1.22 L atm

Total work W2 = -4.90 L atm + ( -1.22 L atm) = -6.125 L atm

But we must flip the sign due to the negative sign in the formula -(P)(change in Volume). So we will get a final answer of a positive 6.125L atm for the work done for the two-step process.

3) Now we calculate the difference between the q values in liter-atmospheres.

This is a rather simple step when given the formulas. q2 - q1 = w1 - w2.

7.35 - 6.125 = 1.225 L atm

1.225 L atm (101.3) = 124.1 J