Answer:
Given that
T₁=16⁰C
V₁=0.0042 m³
P=101.3 KPa
1)
Assuming that gas is ideal gas.
At constant pressure
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
V₂=1.09 V₁
T₁=16⁰C=289 K
By putting the values
[tex]\dfrac{1.09V_1}{1V_1}=\dfrac{T_2}{289}[/tex]
T₂=315.01 K
T₂=42.01⁰C
2)
The internal energy for air
ΔU= m Cv ΔT
P V = m RT ( ideal gas equation)
[tex]m=\dfrac{101.3\times 0.0042}{0.287\times 289}\ kg[/tex]
m=0.0051 kg
Cv for air = 0.71 KJ/kg.k
By putting the values
ΔU= m Cv ΔT
[tex]\Delta U=0.0051\times 0.71\times (42.01-16)\ KJ[/tex]
ΔU=0.094 KJ
3)
V₁=0.0042 m³
V₂=1.09 V₁
V₂=0.0045 m³
Work W
W= P.ΔV
W= 101.3 x (0.0045-0.0042)
W=0.038 KJ
4)
We know that
From first law of thermodynamics
Q= ΔU+ W
By putting the value
Q= 0.038+0.094
Q=0.132 KJ
