The temperature of 3.0 kg of krypton gas is raised from -20^∘ C to 80^∘ C. (a) If this is done a

The temperature of 3.0 kg of krypton gas is raised from...

Key Concepts

Work in Thermodynamic Processes

Work, in the context of thermodynamics, is often performed by the system when it expands or is compressed. In constant pressure processes, the work done is typically calculated as the product of the external pressure and the change in volume, while in constant volume processes, the work is zero.

Constant Pressure Process

During a constant pressure process, the pressure remains unchanged while the volume is allowed to change, resulting in work being done by or on the system due to expansion or compression. This work must be accounted for in addition to the change in internal energy when applying the First Law of Thermodynamics.

Specific Heat Capacities (c_v and c_p)

Specific heat capacities, c_v at constant volume and c_p at constant pressure, measure the amount of heat required to raise the temperature of a unit mass of a substance by one degree under constant volume or pressure conditions, respectively. They play a key role in calculating the heat transfer in thermodynamic processes for gases.

First Law of Thermodynamics

This principle states that the total energy change in a system is equal to the heat added to the system minus the work done by the system. It provides the fundamental relationship among heat, work, and internal energy, and is crucial for analyzing different thermodynamic processes, whether at constant volume or constant pressure.

Constant Volume Process

In a constant volume process, the volume of the system remains fixed, which means that no work is done by the system since work is proportional to changes in volume. In such a process, any heat added to the system directly increases the internal energy.

Transcript

00:01 We have to find the value of work done and internal energy for constant pressure and volume solution this is a part for constant volume for constant volume and some values are given that three gram 3 gram krypton so 3 0 .000 of kryptum that means mass is equal to 3 gram of krypton and the value of temperature change temperature change is equal to from minus 20 degrees celsius to 80 degrees celsius so first at constant volume at this is one at constant volume we know that delta v is 0 and the work done the value of work done is equal to work done w is equal to p delta v so p is multiply by 0 so this is work done is equal to 0 so the first part is said that the value the value of work done is 0 so now we can say that the value of work done is equal to 0 and the second one is the change in internal energy so second the change in internal energy we know that delta q is equal to delta u plus w and this is why first law of thermodynamics so here work done is zero so delta u should be equal to delta u plus zero so this is equal to delta u so the value of change in internal energy is equal to the the value of internal energy is equal to the heat so here we can write that is that and further we know that the value of delta q should be equal to m cb delta t so the value of m is equal to 3 ,000 and cb is given by 0 .0357 and the temperature should be equal to 80 minus of minus 20 by solving this by get that is the value is that 10710 calories in other form we can write that in other form can write that the value is 10 .71 kilo calorie and approximate we can return that 11 kilo calorie so now the value of delta q should be equal to delta u is equal to 10 .71 kilo calorie and 10 .71 is multiply by 4 .184 and we get this in form of jule so the value of delta u should be equal to 44 .81 kilojou in other form delta u should be equal to 45 kilo jule so the next one is the value of internal energy is equal to 45 kilojoules similar we have to find the value of b part but at constant pressure so the second part is that this is b part at constant pressure.

04:23 We know that delta q should be equal to mcp delta t at constant pressure and the value is given that 3 ,000 is multiplied by 0 .0 .05 is multiplied by 80 minus of minus 20.

04:39 By solving this we get the value of delta q should be equal to 17 ,850 calories otherwise 17 ,000.

04:48 Point eight five kilo calorie so the value of delta q should be is equal to 18 kilo calorie so the value of heat should be equal to 18 kilo calories and the next one is that we can write this is that work done so the value of work done is is equal to minus p delta v so this is we can write that in other form p b2 minus b1 what is this is positive so we can write this n r t2 minus t1 so by solving this p n is equal to 3 000 upon 84 and r should be equal to 3 .1 sorry 3 .8 3 3131 8 314 3 and the value of t is equal to 80 minus minus 20 so by solving this we get this is the 29692 jule in other form 29 .69 kilojoule so the value of work done should be equal to 30 kilojoult approximately so this is the value of work done and the last one is we have to find that is the internal energy so the value of internal energy is that we know that delta u is equal to delta q minus w...

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