The experiment described in Exercise 18.124 is repeated, but...

The experiment described in Exercise 18.124 is repeated, but this time a weight, which exerts a pressure of 2 atm, is placed on the piston. When the gas expands, its pressure drops to this 2 atm pressure. Then the weight is removed and the gas is allowed to expand again to a final pressure of 1 atm. Throughout both expansions the temperature of the apparatus was held at a constant $25^{\circ} \mathrm{C}$. Calculate the amount of work performed by the gas in each step. How does the combined total amount of work in this two-step expansion compare with the amount of work done by the gas in the one-step expansion described in Exercise $18.124 ?$ How can even more work be obtained by the expansion of the gas?

Key Concepts

Role of External Pressure in Gas Expansion

External pressure, such as that exerted by a weight on a piston, directly controls the path of the gas expansion. When the gas expands against a controlled pressure, the work done is affected by whether the process is maintained at a constant external pressure or allowed to adjust smoothly. Understanding how external constraints influence the pressure-volume work is key to determining how modifications to the process — like multi-step expansions or near-reversible paths — can be used to optimize work output.

Reversible versus Irreversible Processes

A reversible process is an idealized, quasi-static process in which the system remains in thermodynamic equilibrium at every stage. Such a process maximizes the work extracted from or done on the system. In contrast, irreversible processes, which occur faster or under imposed constraints (like a constant external pressure), yield less work due to finite deviations from equilibrium. This distinction is critical when comparing different expansion paths and assessing strategies for work maximization.

Isothermal Process

An isothermal process is one in which the temperature of the system remains constant throughout the expansion or compression. For an ideal gas, this means that the product of pressure and volume remains constant (PV = constant) according to Boyle’s Law, and the work can be calculated using the integral form W = nRT ln(Vf/Vi). This concept is foundational in understanding how energy is transferred as work without any change in internal energy.

Calculation of Work via Pressure-Volume Integration

The work done by a gas is determined by the integral of pressure with respect to volume change, W = ? P dV. In isothermal processes, the pressure can be expressed in terms of volume using the ideal gas law, leading to the logarithmic relationship. This method of calculation underpins the quantitative analysis of work in both reversible and irreversible processes.

Transcript

00:01 At constant temperature, we have the relationship p1 v1 is equal to p2 v2.

00:14 And you calculate the work, work is equal to negative p is the pressure, delta v is a change in volume.

00:24 So part a, we're as to calculate the work done when the piston goes from 5 .5 liters to 10 .5 liters.

00:33 So we'll first calculate the pressure.

00:37 So initial pressure is three atmospheres and this is at a volume of 5 .5 liters.

00:46 And the final pressure is there unknown and this would be at a volume of 10 and a half liters.

00:52 And so for p2 the final pressure, this works up to 1 .571 atmospheres.

00:58 And then calculate the work done.

01:02 There's a good negative p delta v.

01:04 Negative p pressure is 1 .571 atmospheres and the change in volume is 10 .5 liters minus 5 and a half liters and this will give us a value of negative points 7 .855 um sorry 7 .855 um sorry 7 .855 uh...

01:32 Leader atmosphere and the sign here is negative so i hear work done is the expansion of the gas as it goes from 5 .5 liters to 10 .5 liters.

01:56 So the system works on the surroundings and loses some heat.

02:19 Therefore, it has a negative sign.

02:26 For part b, let's calculate the work done for a two -step process.

02:38 We're going to go from an initial volume of 5 .5 liters to, from step one here, it would go to 7 liters.

02:51 And then in step 2 here, we'd reach 10 .5 liters.

02:56 So we still have the expansion of a gas, but occurring in two steps.

03:01 So for step one, and let's calculate the pressure change, starting from three atmospheres, 5 .5 liters initially.

03:21 P2 is a pressure at after the first step, which is at 7 liters.

03:27 P2 works out to 2 .357 atmospheres.

03:33 And the work done equals negative p delta v, and delta v here is 7.

03:45 Liters minus three or five and a half liters...