# Thermodynamics Cycle: Gas Temperature and Work Calculation | Homework Solution

• kelvin56484984
In summary, the gas is heated at constant pressure until the volume is 4.4L at point B, cooled at constant volume until the pressure decreases to 1.2 atm, and undergoes an isothermal compression back to point A. The temperature at points A, B, and C are 64.5K, 129K, and 64.5K respectively. The work done on the gas, heat transferred to the gas, and change for each process is 535J, 457J, and 992J, respectively. For the whole cycle, the work is 653J, the heat is 54J, and the change in internal energy is 707J. The number of moles,
kelvin56484984

## Homework Statement

I.The gas is heated at constant pressure until the volume is 4.4L at point B
II. The gas is cooled at constant volume until the pressure decreases to 1.2 atm
III. The gas undergoes an isothermal compression back to point A

a) Find the temperature at Point A,B,C
b) Find the work done on the gas ,the heat transferred to the gas and the change for each process
c) Find W,Q and delta U for the whole cycle

## Homework Equations

[/B]
delta U=Q+W
PV=nRT
1 atm=1.013x10^5 N/m^2
Cp=5/3R+R

## The Attempt at a Solution

a)
pv=nRT
Tb=PbVb/nR
Tb=2.4*4.4/(1*8.31) * 0.001* 1.013*10^5
=129K
Ta=129K/2
=64.5K
Tc=129K/2
=64.5K

b)
Wab=pv
=2.4*(4.4-2.2)* 1.013x10^5 *0.001
=535J
Qab=n*Cp*deltaT
=0.32* 8/3*R *(129-64.5K)
=457J
U=Q+W
delta Uab=457+535
=992J

Wbc=0
Since the volume keep unchange
Qbc=n*Cv*deltaT
=0.32*5/3*R*(64.5K-129K)
=-285J
U=Q+W
U=-285J

U=0
Since it is isothermic process
Wca=nRTc *ln(Vc/Va)
=0.32*8.31*64.5K*ln(4.4/2.2)
=118
Q+W=0
Q=-118J

c)
W=Wab+Wbc+Wca
=535+0+118
=653J
Q=Qab+Qbc+Qca
=457-285-118
=54J
deltaU=992-285+0
=707J

#### Attachments

• cycle.JPG
5.4 KB · Views: 407
1. How many moles of gas are there? You seem to be using 1 and .32 in different places. There has to be more information provided in order to determine this.
2. What is the ##\Delta U## for one complete cycle? (hint: is U a state function?).
3. What kind of gas is this? (e.g monatomic, diatomic etc. Are you given this or given the Cv or Cp?).

AM

## What is the Thermodynamics Cycle?

The Thermodynamics Cycle is a sequence of thermodynamic processes that involve the transfer of heat and work between a system and its surroundings. It is used to study the behavior of a gas and its changing temperature as it goes through various stages.

## What is gas temperature?

Gas temperature is a measure of the average kinetic energy of gas molecules. It is directly related to the speed at which the molecules are moving and is typically measured in units of Kelvin (K) or degrees Celsius (°C).

## How is work calculated in a thermodynamics cycle?

In a thermodynamics cycle, work is calculated by finding the area under the curve on a pressure-volume (PV) diagram. This area represents the work done by the gas as it expands or contracts during a process.

## What is the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that the total energy of a closed system remains constant.

## What factors affect the gas temperature in a thermodynamics cycle?

The gas temperature in a thermodynamics cycle can be affected by several factors such as the amount of heat added or removed, the work done by the gas, and the type of process (isothermal, adiabatic, etc.). The specific gas properties, such as specific heat capacity, also play a role in determining the gas temperature.

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