A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in the p V -diagra

step 1 So this is question 20 .34. So we use the ideal gas law to calculate P and V for each state. We

A heat engine takes 0.350 mol of a diatomic ideal gas...

A heat engine takes $0.350 \mathrm{~mol}$ of a diatomic ideal gas around the cycle shown in the $p V$ -diagram of Fig. P20.34. Process $1 \rightarrow 2$ is at constant volume, process $2 \rightarrow 3$ is adiabatic, and process $3 \rightarrow 1$ is at a constant pressure of 1.00 atm. The value of $\gamma$ for this gas is 1.40 .1 .0 (a) Find the pressure and volume at points $1,2,$ and 3 . (b) Calculate $Q, W,$ and $\Delta U$ for each of the three processes. (c) Find the net work done by the gas in the cycle. (d) Fing the net heat flow into the engine in one cycle. (e) What is the therma efficiency of the engine? How does this compare to the efficiency of a Carnot-cycle engine operating between the same minimum and maximum temperatures $T_{1}$ and $T_{2}$ ?

Key Concepts

Thermal Efficiency in Heat Engines

Thermal efficiency measures the fraction of heat input that is converted into work by the engine. It is defined as the net work output divided by the total heat input. Efficiency analysis is crucial for comparing real engines with the theoretical maximum efficiency given by the Carnot cycle.

First Law of Thermodynamics

This law states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system (?U = Q - W). It is a fundamental principle for performing energy balance calculations during each step of a thermodynamic cycle.

Heat Transfer

Heat transfer refers to the energy exchanged between a system and its surroundings due to temperature differences. In thermodynamic cycles, understanding heat transfer during each process is essential for calculating the overall energy balance.

Carnot Efficiency

Carnot efficiency represents the maximum possible efficiency of a heat engine operating between two temperature reservoirs (T_high and T_low). It serves as an important benchmark for evaluating the performance of real engines.

Work in Thermodynamic Processes

Work is done by the gas when it expands or is compressed, and it can be calculated as the area under (or above) the process curve on a p-V diagram. Different processes (isobaric, isochoric, and adiabatic) have specific formulas for the work done.

Ideal Gas Law

This law (PV = nRT) relates the pressure, volume, and temperature of an ideal gas and is fundamental in determining the state points of a gas during thermodynamic processes. It is crucial for analyzing cycles involving ideal gases and understanding how these state variables change during different processes.

Adiabatic Process

An adiabatic process occurs without any heat exchange with the surroundings. In such a process, any work done results in a change in the internal energy, leading to a change in temperature that depends on the heat capacity ratio (gamma).

Processes at Constant Volume (Isochoric Process)

In an isochoric process, the volume remains constant; hence, no work is done by or on the gas. Any heat added or removed from the system directly changes its internal energy.

p-V Diagram Analysis

A p-V diagram graphically represents the relationship between pressure and volume during thermodynamic processes. It is essential for visualizing cycles, identifying the types of processes (e.g., constant volume, adiabatic, constant pressure), and calculating work done by or on the gas through areas under the curves.

Processes at Constant Pressure (Isobaric Process)

During an isobaric process, the pressure remains constant, which means that the work done by the gas is directly related to the change in volume. The heat transfer in this process also affects the internal energy and the work done.

Transcript

00:01 So this is question 20 .34.

00:04 So we use the ideal gas law to calculate p and v for each state.

00:09 We use the first law in specific expressions for qw and changing u for each process.

00:16 So to start off with a, for 0 .1, we have p1 equals 1 atm, which is 1 .03 times 10 to the 5th pass scale.

00:33 And then we have pv is equal to nrt.

00:39 So v1 is equal to nrt1 over p1, which is equal to 0 .350 moles times 8 .3145 joules per mole times kelvin, times 300 kelvin, divided by 1 .013 times 10 to the 5th pascels.

01:03 And that leaves you with 8 .62 times 10 to the negative third meters cubed.

01:10 For point two, we have v2 is equal to v1, which is equal to 8 .62 times 10 to the negative third meters cubed.

01:22 So pv is equal to nrt, and nrv constant applies or implies that p1 divided by t1 is equal to p2 over t2.

01:35 So if we have to solve for p2, p2 is equal to p1 divided by t2 divided by t1, and so that is 1 .0 atms.

01:48 Then 600 kelvin divided by 300 kelvin, and that leaves you with 2 .03 times 10 to the 5th pascels.

02:00 For 0 .3, we have p3 is equal to p1, which is equal to 1 atm, which is equal to 1 atm, which is equal to 1atm, which is equal to 1 .013 times 10 to the 5th pascal's.

02:16 So we have pv equals nrt, and so v1 over t1 is equal to v3 over t3.

02:25 So if we have to solve for v3, v3, v3 is equal to v1 times t3 divided by t1, and so that would be 8 .62 times 10 to negative 3 meters cubed times 492 kelvin, divided by 300 kelvin and that leaves you with 14 .1 times 10 to negative third meters cubed.

02:54 For part b, process 1 to 2 would be q equals n cv change in t, which would be 0 .350 moles times 20 .79 joules and then 600 minus 300...

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