A sample of an ideal gas at 15.0 atm and 10.0 L is allowed to expand against a constant extern

A sample of an ideal gas at 15.0 atm and 10.0 L is allowed...

A sample of an ideal gas at $15.0 \mathrm{~atm}$ and $10.0 \mathrm{~L}$ is allowed to expand against a constant external pressure of $2.00 \mathrm{~atm}$ at a constant temperature. Calculate the work in units of $\mathrm{kJ}$ for the gas expansion. (Hint: Boyle's law applies.)

Key Concepts

Isothermal Process

An isothermal process is one in which the temperature of the system remains constant throughout the process. For an ideal gas, this means that the product of pressure and volume remains constant, and any work done by or on the system is exactly balanced by heat flow into or out of the system, as expressed by Boyle's law.

Boyle's Law

Boyle's law states that for a fixed amount of an ideal gas at constant temperature, the pressure is inversely proportional to the volume. This relationship (P?V? = P?V?) is crucial for determining the final volume of the gas when it expands or compresses under constant temperature conditions.

Work Done by Gas Expansion

The work done by a gas during expansion is given by the integral of the external pressure over the change in volume. In cases where the external pressure is constant, the work can be simply calculated as w = -P_ext ?V. The negative sign indicates that the system does work on the surroundings, meaning energy is leaving the system.

Unit Conversion

Converting work from one set of units to another is an essential skill in thermodynamics. When work is calculated in atm·L, it is typically converted to joules or kilojoules using the conversion factor that 1 atm·L is approximately equal to 101.325 joules. This conversion is vital for ensuring consistency and correctness in energy calculations.

Transcript

00:01 So let's start this question off by writing down what we already know.

00:04 So our ideal gas is held at a pressure of 15 atmospheres and at a volume of 10 liters.

00:15 We know there is an external pressure of two atmospheres.

00:25 We can be able to p1 and p2, respectively.

00:28 So we want to find how much work is being done when it comes to gas expansion.

00:35 So we're seeing here we have two pressures and a volume.

00:39 And as the question suggests, you can use the boyle's law.

00:44 And boyle's law is written as p1, v1 is equal to p2, v, where these two sets of values are proportional to each other.

01:04 So we can actually just plug these numbers in to this equation.

01:13 So p1 we know is 15 atmospheres, v1 is 10 liters.

01:19 And we don't know how much this gas is going to be expanding to, so v2 is going to be left blank.

01:35 And then we just rearrange the equation, and we calculate, and we get 75 liters.

02:02 Okay, so we have the volume towards the final volume of this gase system.

02:07 So in order to find the work being done, we're going to be using this equation, where the amount of work, or lower case w, is equal to the negative of external pressure multiplied by the final volume minus the initial volume.

02:34 So we know the external pressure is two atmospheres, so we'll just write two, and we know these two values already, so we're going to plug that in.

02:52 So it's 2 times 65, so that's 130 liters atmospheres.

03:01 But this presents a problem because we have two units here, but we want a little bit.

03:07 Jules.

03:09 So how do we get it? well, since we're dealing with a gas here, we can use gas constants.

03:16 We can use r.

03:27 And we know that r has different values depending on what units we're using.

03:33 And in this case leaders and atmospheres belong to one set of the gas constant...

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