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An ideal monatomic gas is contained in a vessel of constant volume 0.230 m3. The initial temperature and pressure of the gas are 300 K and 5.00 atm, respectively. The goal of this problem is to find the temperature and pressure of the gas after 23.0 kJ of thermal energy is supplied to the gas.

(a) Use the ideal gas law and initial conditions to calculate the number of moles of gas in the vessel.
mol
(b) Find the specific heat of the gas.
J/K
(c) What is the work done by the gas during this process?
kJ
(d) Use the first law of thermodynamics to find the change in internal energy of the gas.
kJ
(e) Find the change in temperature of the gas.
K
(f) Calculate the final temperature of the gas.
K
(g) Use the ideal gas expression to find the final pressure of the gas.
atm

Respuesta :

Answer:

a)n= 46 .69 moles

b)[tex]C_v=12.47\ J/mol.K[/tex]

c)W = 0

d)ΔU = 23 KJ

e)ΔT= 39.05 K

g)T₂= 339.05 K

P₂ = 5.65 atm

Explanation:

Given that

V₁=V₂ = 0.23 m³ = 230  L

T₁ = 300 K

P₁ = 5 atm        ( 1 atm =100 KPa)

Q = 23 KJ

The ideal gas equation

P V = n R T

P=Pressure  ,  V=Volume  ,  n= Number of moles  , T=Temperature

R=gas constant = 0.0821 atm.L/K.mol

P₁ V₁ = n R T₁

[tex]n=\dfrac{P_1V_1}{RT_1}[/tex]

[tex]n=\dfrac{5\times 230}{0.0821\times 300}[/tex]

n= 46 .69 moles

Specific heat ;

The specific heat of the gas Cv

[tex]C_v=\dfrac{3}{2} R[/tex]

R=8.314 J / mol·K

[tex]C_v=\dfrac{3}{2} \times 8.314\ J /mol.K[/tex]

[tex]C_v=12.47\ J/mol.K[/tex]

We know that work done given as

W = P ΔV

ΔV =Change in volume ,P =pressure

here ΔV = 0

W = 0

First law of thermodynamics

Q= ΔU +W

Q= ΔU = 23 KJ

ΔU =Change in the internal energy

Q = n Cv ΔT

23 000 = 46.69 x 12.47 ΔT

ΔT= 39.05 K

Change in the temperature =39.05 K

T₂- 300 = 39.05 K

T₂= 339.05 K

[tex]P_2=\dfrac{T_2}{T_1}P_1[/tex]

[tex]P_2=\dfrac{339.05}{300}\times 5[/tex]

P₂ = 5.65 atm

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