Respuesta :
Answer:
Explanation:
A ) At constant volume :
ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .
For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J
ΔEint = .7 x 12.45 x ( 564 - 300 )
= 2300.76 J .
W₁ = 0 because volume is constant so work done by gas is zero .
Q₁ = ΔEint = 2300.76 J
B )
At constant pressure
Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .
For monoatomic gas ,
Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J
Q₂ = .7 x 20.75 x 264 J
= 3834.6 J
W₂ = work done by gas
= PΔV = nRΔT
= .8 x 8.3 x 264
= 1752.96 J
ΔEint = Q₂ - W₂
= 3834.6 - 1752.96
= 2081.64 J.
ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as
a)
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Q2= 3834.6 J
W2= 1752.96 J
dE2= 2081.64 J.
What is ΔEint, Q1 and W1 for the process at constant volume and pressure?
a)
Generally, the equation for the Constant Volume is mathematically given as
dE = n Cv x dT
Where
Cv = 3/2
R = 1.5 x 8.3 J
R= 12.45 J
Therefore
dE = 0.7 x 12.45 x ( 564 - 300 )
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Generally
Q2 = n Cp dT
Where
Cp = 5/2
R = 2.5 x 8.3 J
R= 20.75 J
Hemce
Q2 = 0.7 x 20.75 x 264 J
Q2= 3834.6 J
For Work done
W=PdV
W= nRdT
Therefore
W= 0.8 x 8.3 x 264
W2= 1752.96 J
Hence
dE = Q₂ - W₂
dE= 3834.6 - 1752.96
dE2= 2081.64 J.
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