Help Termodynamic problem getting grade tommorow

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SUMMARY

The forum discussion centers on solving a thermodynamics problem involving the isobaric expansion of hydrogen gas. The key equations discussed include the ideal gas law (PV = nRT) and work done during expansion (W = P × ΔV). The user seeks to determine the mechanical work (L), heat absorbed (Q), and change in internal energy (ΔU) given specific parameters: mass (m = 2g), initial temperature (t1 = 300K), and expansion factor (k = 3). The final answers provided are W = 4.99 kJ, Q = 17 kJ, and ΔU = 12.47 kJ.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of isobaric processes in thermodynamics
  • Familiarity with mechanical work calculations (W = F × d)
  • Basic concepts of internal energy and heat transfer
NEXT STEPS
  • Study the derivation and applications of the ideal gas law (PV = nRT)
  • Learn about isobaric and isochoric processes in thermodynamics
  • Explore the relationship between work, heat, and internal energy in thermodynamic systems
  • Investigate the Carnot cycle and its efficiency in thermodynamic processes
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone preparing for exams in mechanical or physical sciences will benefit from this discussion.

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Homework Statement


The volume of a m (mass) hydrogen quantity,which wear initially at the temperature t1, increases k times by isobar dilatation.Find: a) L= ? b) the heath Q absorbed. c)Δ U =?

Homework Equations


How to solve it?




The Attempt at a Solution



pv/t=p2v2/t2 ?
L= p * Δ V ?
I don't know... and I am getting grade tomorrow on it...
 
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What is the meaning of L?
 
L- mecanic force
idk how the say it in english

its the mecanic stuff that in Mecanic phisics is L=f*d and the values are in joules
 
Fxd is called WORK.

Work is defined as the ...
 
yes i found on wiki english its Work :) but we note it whit L
so how i do the problem?
 
W= p x V??
can you tell me what to do?
 
W = force x distance moved

But Pressure P = Force / Area

therefore F = P x A

Hence W = F x d

= P x A x d

= P x change in Volume
 
hence your equation for W is correct since the pressure is constant.
 
what A means
and =P x delta Volume is the equation to fiend W right?
also how do i fiend the pressure? whit p1/t1=p2/t2?
 
  • #10
A means Area.

Use pV = nRT which becomes p1V1/T1 = p2V2/T2 since we assume the number of moles in gas, n, is constant.

But since the process is isobaric the pressure is constant so that the last equation becomes

V1/T1= V2/T2.
 
  • #11
you said earlier that W= P x change in Volume Right?
so what i do whit v1/t1=v2/t2? as i see nothing that can help in the above
 
  • #12
W = P x change in volume
= P x kV1

Other equation can be used to find final temperature required to find the heat given.
 
  • #13
grzz said:
W = P x change in volume
= P x kV1

Other equation can be used to find final temperature required to find the heat given.

and how do i find the p?
 
  • #14
What physical constants are given numerical values in this problem?
 
  • #15
m= 2g
t1=27 celsius=300 Kelvin
k=3
 
  • #16
Is initial volume given a numerical value?
 
  • #17
It say :The volume of a quantity of Hydrogen of mass m , found initial at the temperature t1, rises of K times in izobare dilatation.Fiend the W ,Q absorbed and the Delta u (Intern energy)
m=2g t1=27 Celsius and k=2

For b i think is W=Delta U + Q i think
 
  • #18
Since k is positive there is an expansion, so work is done BY the system and so heat must enter the system.

Hence

change in U = heat in - work by system.
 
  • #19
grzz said:
Since k is positive there is an expansion, so work is done BY the system and so heat must enter the system.

Hence

change in U = heat in - work by system.

yes bur for W what i do? how i fiend the pressure?
 
  • #20
I think that some other physical quantity must be given a numerical value.
 
  • #21
we got to exprime V2=V1 x K i think as it say that the second volume is rising by K times
 
  • #22
PV = nRT

So p = nRT/V
= 1 x R x 300/V

So we have to know R and V to find P.
 
  • #23
grzz said:
PV = nRT

So p = nRT/V
= 1 x R x 300/V

So we have to know R and V to find P.

R has the value 8.314 J·K−1·mol−1.
 
  • #24
And numerical value for initial volume V?
 
  • #25
grzz said:
And numerical value for initial volume V?

i don't know, doesn't mention but here are the answers W= 4,99 kJ Q = 17 Kj Δ U= 12,47Kj
 
  • #26
but in izobaric process the Pressure is constant so we don't use it in equations :)
this helps?
 
  • #27
To find W one NEEDS pressure.
 
  • #28
Try this problem:In a box closed by a piston is a quantity of ideal gas characterized whit the parameters p1,V1,T1.The gass is put to this processes:
1. izocore heating up to p1= 1,5p1
2.izobar uncompression up to V3= 2 V1
3. izocore cooling up to p4=p1
4. izobare comprimation up to initial state.
a) Represent graphic in the coordinates (p1,V)(p,T) and (V,T) the processes suffered.
b)Calculate the temperature of the gass in the state 3,T3, knowing that T1=300K
c)Calculate the randament of a Carnot cicle who would function bettwhen the extreme temperatures touched in this succesion of termodynamic processes
 
  • #29
re 2nd problem:

We are going to use PV = nRT with n and R being constant.

So PV/T is constant.

So P1V1/T1 = P3V3/T3.
 

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