One mole of nitrogen gas at 301 K occupies a volume of 12.5 L in the frictionles

Question

One mole of nitrogen gas at 301 K occupies a volume of 12.5 L in the frictionless piston/cylinder arrangement shown. The mass of the piston may be assumed to be negligible. The crosssectional area of the piston face is 300 cm2 , and the pressure of the external atmosphere is 100 kPa. When a 600 kg weight is placed on the piston, the piston lowers by 13.49 cm. The process may be assumed to be isothermal, and nitrogen may be assumed to behave as an ideal gas. The acceleration due to gravity is 9.80 m s–2 .

What is the minimum work required to carry out this compression?

Please write steps clearly.

Explanation / Answer

V1 = initial volume = 12.5 L = 0.0125 m3

P1 = initial pressure = atmospheric pressue = 105 Pa

A = area of piston = 300 x 10-4 m2

P2 = final pressure = P1 + mg/A = 105 + (600) (9.8)/(300 x 10-4) = 2.96 x 105 Pa

V2 = final Volume = ?

using : P1 V1 = P2 V2

( 105 ) (0.0125) = (2.96 x 105) V2

V2 = 0.0042 m3

Change in Volume = V' = V1 - V2 = 0.0125 - 0.0042 = 0.0083

work done is given as

W = - P2 V' = - (2.96 x 105) (0.0083) = - 2456.8 J

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.