A gas is confined to a container with a massless piston at the top. (Part C figure) 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 2.45 L. When the external pressure is increased to 2.50atm, the gas further compresses from 2.45 to 1.96 L.

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the gas, decreasing its volume from 4.90 to 1.96 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?

124 J

371.175J

123J

Well, it seems like we've got a gas that's feeling a little bit compressed! So let's see what's going on here. In the two-step process, the gas goes from 4.90 to 2.45 L and then to 1.96 L, while in the one-step process, it goes straight from 4.90 to 1.96 L.

Now, the final temperature is the same for both processes, so that means we can assume that the gas is experiencing an isothermal process. This means that the change in internal energy (ΔU) is equal to zero, because ΔU = q + w and w (the work done on or by the system) is equal to zero for an isothermal process.

Since the change in internal energy is zero, that means that the heat transfer for both processes (q) must be the same. So, the difference between q for the two-step process and q for the one-step process is zero joules!

It looks like this gas is feeling a little "deflated" when it comes to differences in heat transfer. But hey, at least it's keeping its cool temperature-wise!

To find the difference between q (heat) for the two-step process and q for the one-step process, we need to calculate the heat for each process separately and then find the difference between them.

Let's start by calculating the heat (q) for the two-step process. In this process, the gas undergoes two separate compressions. We know the initial and final volumes, as well as the external pressure. However, we need additional information to calculate the heat. Specifically, we need either the pressure or the number of moles of gas.

Since the experiment is separate and doesn't provide the pressure or the number of moles, we will have to use the ideal gas law to determine the missing information. The ideal gas law is given by:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Since the initial conditions of volume and pressure are the same in both experiments, we can use the ideal gas law to calculate the number of moles for the two-step process using the first set of conditions:

P1V1 = nRT

Rearranging the equation, we have:

n = P1V1 / RT

Now we have the number of moles for the two-step process. We can use this information to calculate the heat (q).

q = nCΔT

where C is the molar heat capacity and ΔT is the change in temperature.

To compare and find the difference between q for the two-step process and q for the one-step process, we need to calculate the heat for the one-step process using the given volume change and pressure.

Again, we need the number of moles for the one-step process. Since the initial conditions for the one-step process are the same as the final conditions for the two-step process, we can use the number of moles we calculated earlier.

Using the number of moles, we can now calculate the heat (q) for the one-step process.

Finally, to find the difference between the two heats (q values), subtract the heat of the one-step process from the heat of the two-step process: q_two_step - q_one_step.

WRONG