A piston–cylinder device initially contains 0.07m^{3} of nitrogen gas at 1

Chesley

Chesley

Answered question

2021-09-23

A piston–cylinder device initially contains 0.07m3 of nitrogen gas at 130 kPa and 180. The nitrogen is now expanded to a pressure of 80 kPa polytropically with a polytropic exponent whose value is equal to the specific heat ratio (called isentropic expansion). Determine the final temperature and the boundary work done during this process.

Answer & Explanation

brawnyN

brawnyN

Skilled2021-09-24Added 91 answers

Determine the final volume of nitrogen V2.
p1V1k=p2V2k 
V2=(p1p2V1k)1k 
V2=(130kPa80kPa(0.07m3)1.4)11.4 
V2=0.0990m3 
W=p2V2p1V11n 
W=80kPa0.0990m330kPa0.07m311.4 
W=2.95kJ

user_27qwe

user_27qwe

Skilled2023-06-19Added 375 answers

Given:
Initial volume (V1) = 0.07 m3
Initial pressure (P1) = 130 kPa
Initial temperature (T1) = 180°C
Final pressure (P2) = 80 kPa
Polytropic exponent (n) = specific heat ratio (γ)
To find:
Final temperature (T2) and boundary work done (Wboundary)
The process is an isentropic expansion, which means the entropy remains constant (ΔS=0).
Using the ideal gas law, we can write:
PV=mRT
Where:
P is the pressure
V is the volume
m is the mass of the gas
R is the specific gas constant
T is the temperature
Assuming the mass of the nitrogen gas remains constant, we can rewrite the equation as:
P1V1T1=P2V2T2
Since the process is isentropic, we also have:
P1P2=(V2V1)γ
Solving these two equations simultaneously, we can find T2 and V2.
Now, let's solve for V2:
P1P2=(V2V1)γ
Taking the logarithm of both sides:
log(P1P2)=γlog(V2V1)
Solving for V2:
V2V1=(P1P2)1/γ
Substituting the given values:
V20.07=(13080)1/γ
Simplifying further, we have:
V2=0.07×(13080)1/γ
Now, let's solve for T2:
P1V1T1=P2V2T2
Solving for T2:
T2=P2V2T1P1V1
Substituting the given values:
T2=80×0.07×180130×0.07=80×180130
Simplifying further, we have:
T2=14400130
Finally, let's calculate the boundary work done (Wboundary):
Wboundary=P2V2P1V11n
Substituting the given values:
Wboundary=80×0.07130×0.071γ
Simplifying further, we have:
Wboundary=0.07×801301γ
Therefore, the final temperature T2 is 14400130 and the boundary work done Wboundary is 0.07×801301γ.
alenahelenash

alenahelenash

Expert2023-06-19Added 556 answers

Step 1: Calculate the initial temperature using the ideal gas equation.
Given:
V1=0.07m3 (initial volume)
P1=130kPa (initial pressure)
T1=180C=180+273.15K (initial temperature)
We can rearrange the ideal gas equation to solve for T:
T=P·Vm·R
The mass m is unknown, but we can use the ideal gas equation to find it:
PV=mRT
Solving for m gives:
m=PVRT
Substituting this value of m back into the equation for temperature, we have:
T=P·V(PVRT)·R=TP
Plugging in the given values, we get:
T1=180+273.15130K
Step 2: Calculate the final temperature using the polytropic process equation.
Given:
P2=80kPa (final pressure)
n=γ (polytropic exponent, where γ is the specific heat ratio)
Using the polytropic process equation, we can write:
P1·V1n=P2·V2n
Since the initial and final volumes are not given, we can write:
V2V1=(P1P2)1/n
V2=V1·(P1P2)1/n
Now, let's calculate the final volume V2 using the given values:
V2=0.07m3·(13080)1/γ
Step 3: Calculate the final temperature using the ideal gas equation.
Now that we have the final volume V2, we can use the ideal gas equation to calculate the final temperature T2:
T2=P2·V2m·R
Step 4: Calculate the boundary work done during the process.
The boundary work done during the process can be calculated using the formula:
W=V1V2PdV
Since the process is isentropic, we can use the equation:
PVn=constant
To write:
P=constantVn
Substituting this into the work equation, we have:
W=V1V2constantVndV
Simplifying, we get:
W=constantn1(V21nV11n)
Since we already have the values of V1 and V2, we can calculate the work using the given polytropic exponent n=γ.

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