Answer:
[tex]V_2 = 0.125 m^3[/tex]
Work done = = 5 kJ
Explanation:
Given data:
volume of nitrogen [tex] v_1 = 0.08 m^3[/tex]
[tex]P_1 = 150 kPa[/tex]
[tex]T_1 = 200 degree celcius = 473 Kelvin[/tex]
[tex]P_2 = 80 kPa[/tex]
Polytropic exponent n = 1.4
[tex]\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}[/tex]
putting all value
[tex]\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}[/tex]
[tex]\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS[/tex]
polytropic process is given as
[tex]P_1 V_1^n = P_2 V_2^n[/tex]
[tex]150\times 0.08^{1.4} = 80 \times V_2^{1.4}[/tex]
[tex]V_2 = 0.125 m^3[/tex]
work done [tex]= \frac{P_1 V_1 -P_2 V_2}{n-1}[/tex]
[tex]= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}[/tex]
= 5 kJ