Answer:
a)P₂=3236.21 KPa
b)E=3620.26 KPa
Explanation:
Given that
V₁= 1 m³
T₁=40°C
P₁= 340 KPa
V₂=0.2 m³
k= 1.4
We know that for isentropic process
[tex]PV^{k}=C[/tex]
[tex]P_1V_1^{k}=P_2V_2^{k}[/tex]
[tex]340\times 1^{1.4}=P_2\times 0.2^{1.4}[/tex]
P₂=3236.21 KPa
Final pressure of gas P₂=3236.21 KPa
We know that modulus of elasticity
[tex]E=\dfrac{dP}{-\dfrac{dV}{V}}[/tex]
[tex]E=\dfrac{3236.21-340}{-\dfrac{0.2-1}{1}}[/tex]
E=3620.26 KPa