The figure shows a reversible cycle through which 1.00 mole...

Name: The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with pb = 13.0 atm and Vb = 3.00 x 10^-3 m^3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.
Uploaded: 2022-09-25T17:59:38-08:00
Duration: 8 min 9 s
Channel: Ivan Kochetkov
Description: The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with pb = 13.0 atm and Vb = 3.00 x 10^-3 m^3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with pb = 13.0 atm and Vb = 3.00 x 10^-3 m^3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

Transcript

00:01 Hello, here we have to solve the following problem.

00:04 The process bc shown in the screen is adiabatic and pressure at point b is 13 atmospheres.

00:11 Meanwhile, the volume and point b is 3 times 10 power by negative 3 cubic meters.

00:20 In question a, we have to calculate energy added to the gaze as heat.

00:28 This energy equals to the heat.

00:32 Which is added in the process ab.

00:36 This process is isohoric.

00:44 It means that work done by the gase is zero joules.

00:47 And this energy equals to 3 over 2 because the gaze is monotomic times nr, tb minus t .a.

00:56 Or that is also 1 .5 times p bvb minus p .a vb minus p .a.

01:07 Now let's look at the process bc which is adiabatic that's why pv power by gamma is constant and here that is this gamma is 1 .67 because gase is monotomic.

01:25 It means that pb vb power by 1 .67 equals to pcvc power by 1 .67.

01:38 Then, pb over pc, or it's better to write it as pc over pb is vb over vc power by 1 .67.

01:58 And therefore, pa is the same as pc, which is pb times 8 power by 1 .67.

02:17 Let's calculate it.

02:26 That is 32 .2 pb and now we can calculate qab.

02:37 Qab equals to 1 .5 times oh actually i think i've made the typo let me double check it.

02:54 Yeah in in this ratio that is that has to be 1 eighths power by 1 .67 therefore pa is the same as pc.

03:05 Which is p b over 32 .2 so it means that uab is 1 .5 times p b which is 13 times 101 325 pascal over 32 .2 multiplied by so actually here we have to subtract here we have to subtract here we have to subtract pb from pb which is basically 13 times 101 325 pascal we multiply this number by 1 minus 1 over 32 .2 and we multiply it by the volume at point b and point a which is 3 times 10 power by negative 3 cubic meters...

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