A sample of nitrogen gas expands in volume from 2.5 to 5.8 L at constant temperature. Calculate

A sample of nitrogen gas expands in volume from 2.5 to 5.8...

A sample of nitrogen gas expands in volume from $2.5$ to $5.8 \mathrm{~L}$ at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of $1.35$ atm, and (c) against a constant pressure of $6.9$ atm. (See Equation 10.2.)

Key Concepts

Unit Conversion (L·atm to Joules)

In thermodynamics, work may be expressed in different units. When the work calculated in liter-atmospheres (L·atm) needs to be converted to joules (J), a conversion factor is used. One liter-atmosphere is approximately equal to 101.3 joules, enabling the conversion of work calculations to the International System of Units.

Pressure-Volume Work

This concept refers to the work done by or on a system when its volume changes against an external pressure. In thermodynamics, work is calculated using the integral of the external pressure with respect to volume change (w = -? P_ext dV), where the negative sign indicates that work done by the system (expansion) is considered negative under the usual sign convention.

Constant External Pressure Work

When a gas expands or compresses under a constant external pressure, the work done can be conveniently calculated using the formula w = -P_ext*(V_final - V_initial). This simplification is particularly useful in many thermodynamic problems where the external pressure remains fixed during the process.

Free Expansion

Free expansion refers to the process where a gas expands against zero external pressure, such as into a vacuum. Since the external pressure is zero, no work is done by the gas during this type of expansion, illustrating an important case in thermodynamics where the system’s expansion does not result in work output.

Transcript

00:01 So for this question, we're asked to calculate the work done in joules for the expansion of a gas under three different conditions.

00:11 So the information that we're given is that the gas expands from 2 .5 liters to 5 .8 liters and that the temperature is held constant.

00:21 So in order to calculate the work, we can use the equation where work equals negative pressure, times the change in volume.

00:35 So in order to calculate the change in volume, we just take the largest, minus the smallest, you put in your calculator, and the change in volume should be 3 .3 liters.

00:56 All right, so let's start with part a, and we are told that is expanding in a vacuum.

01:04 So when something is in a vacuum, it means that the pressure equals zero because there's no pressure.

01:13 It's in a vacuum, there's no pressure.

01:16 So pressure is going to be zero atmospheres.

01:19 So we just plug everything into the equation.

01:21 So work equals negative pressure, which is zero atmospheres, times the change in volume, which we calculated to be 3 .3 liters.

01:37 Now we're asked to put work into jewels.

01:40 So we just have to do a unit conversion.

01:42 Right now we have the units atmosphere times liters or leaders times atmosphere.

01:47 So we're going to do a unit conversion where 101 .3 joules is equal to one liter times atmosphere.

01:57 So the units cancel till we're left with jewels.

02:02 And then zero times anything is zero.

02:04 So the work for these conditions and events.

02:09 Vacuum is zero joules.

02:13 So for part b, we are told that the pressure is constant at 1 .35 atmospheres...

Please give Ace some feedback