Answer:
Work done in this process = 4053 J
Explanation:
Mass of the gas = 0.092 kg
Pressure is constant = 1 atm = 101325 pa
Initial temperature [tex]T_{1}[/tex] = 200 K
Final temperature [tex]T_{2}[/tex] = 200 - 85 = 115 K
Gas constant for nitrogen = 297 [tex]\frac{J}{kg k}[/tex]
When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.
⇒ V ∝ T
⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{T_{2} }{T_{1} }[/tex] ------------ ( 1 )
From ideal gas equation [tex]P_{1}[/tex] [tex]V_{1}[/tex] = m R [tex]T_{1}[/tex] ------ (2)
⇒ 101325 × [tex]V_{1}[/tex] = 0.092 × 297 × 200
⇒ [tex]V_{1}[/tex] = 0.054 [tex]m^{3}[/tex]
This is the volume at initial condition.
From equation 1
⇒ [tex]\frac{V_{2} }{0.054}[/tex] = [tex]\frac{200}{115}[/tex]
⇒ [tex]V_{2}[/tex] = 0.094 [tex]m^{3}[/tex]
This is the volume at final condition.
Thus the work done is given by W = P [[tex]V_{2}[/tex] - [tex]V_{1}[/tex] ]
⇒ W = 101325 × [ 0.094 - 0.054]
⇒ W = 4053 J
This is the work done in that process.