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 = 7.30 atm and Vb = 1.30 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: 2023-05-14T17:36:48-08:00
Duration: 6 min 48 s
Channel: Madhur L
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 = 7.30 atm and Vb = 1.30 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 = 7.30 atm and Vb = 1.30 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 Hi students, here in the first part of the question we have to find out energy added to the gas as heat.

00:09 So, energy added as heat during the portion of the process from a to b, this portion occurs at constant bb, isn't it? so, here q in is equal to n into cv delta t.

00:31 Okay, so the gas is a monoatomic ideal gas, monoatomic ideal gas, so cv is equal to 3r by 2 and the ideal gas law gives delta t is equal to 1 by nr into pvvb minus pava.

01:03 Okay, so 1 by nr pv minus pa because vb and va are equal to pv.

01:17 Okay, so this implies thus q in is equal to 3 by 2, 3 by 2 pv minus pa into vb into bb.

01:39 Okay, so pv is given, we need to find out what is pa.

01:48 Okay, so we need to find va and pa, pa and pa is same as pc, isn't it? pa is same as pc, so this will be pa and pa and pc are same.

02:05 So, for the c and b it's an adiabatic process, isn't it? so, these are connected by adiabatic process, thus we can write pcvc gamma is equal to v is to gamma is equal to pvvb raise to gamma.

02:24 Okay, so pa is equal to pc, so that's equal to bb by vc raise to gamma.

02:40 Okay, into pb, so pa is equal to 1 by 8 all raise to 5 by 3 into 1 .013 into 10 raise to 6, 6, so that's equal to 3 .167 into 10 raise to 4, so this is pa.

03:12 Okay, therefore q in is equal to 3 divided by 2 into 1 .013 into 10 raise to 6 minus 3 .167 into 10 raise to 4 into, into 1 into 10 raise to minus 3, so that's equal to 1 .47 into 10 raise to 3 joule.

03:45 Okay, therefore energy added as heat is 1 .47 into 10 raise to 3 joule.

03:52 For the next part of the question, we have to find out the energy leaves the gas as heat.

03:58 So, energy leaves the gas as heat during the portion of the process from c to a.

04:06 So, this is a constant pressure process.

04:10 So, q out can be ncp delta t, so that's equal to 5 by 2 pa va minus pc vc, so that's equal to 5 by 2 pa into constant pressure, so pa into va minus vc...

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