A piston-cylinder device contains 0.05 m3 of gas initially at 200kpa. at this state, a linear spring that has a spring constant of 150 kn/m is touching the piston but exerting no force on it. now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. if the cross-sectional area of the piston is 0.25 m2, determine (a) the final pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work against the spring to compress it.
Answers
Given : Volume is 0.05 m³, k is 150 kN/m and area is 0.25 m² and pressure innitial is 200 kPa
To find: (a) the final pressure inside the cylinder,
(b) the total work done by the gas, and
(c) the fraction of this work against the spring to compress it.
Solution:
- (a) Now we have provided with volume, area and pressure, so the final pressure will be:
- Let the final pressure be P(f) and the innitial pressure be P(1).
- Now P(f) = P(1) + P(spring)
- But P(spring) = F(spring) / A = kx / A = k(ΔV/A) / A
k × ( (0.1 - 0.05) m³ / 0.25 m² ) / 0.25 m²
50 kN/m × 0.2 m / 0.25 m²
30 kN / 0.25 m²
120 kPa
- So, P(f) = P(1) + P(spring)
- P(f) = 200 kPa + 120 kPa
- P(f) = 320 kPa
- (b). Now for total work done, let work done by piston be W1 and work done by spring be W2 and total work done be W.
- So, W1 + W2 = W
P(1)(V2 - V1) + k/2 x (x2² - x1²)
200 x (0.1 - 0.05) x (1 kJ/ 1 k Pa m³) + 1/2 x 150 x (0.2² - 0²) x (1 kJ/ 1 k Pa m³)
- After calculation, we get:
- W = 10 kJ + 3 kJ = 13 kJ
- (c). For work against the spring, let it be W(s). So:
W(s) = W(2) = k/2 x (x2² - x1²)
W(s) = 1/2 x 150 x (0.2² - 0²) x (1 kJ/ 1 k Pa m³)
W(s) = 3 kJ
Answer:
So, the final pressure inside the cylinder is 320 kPa, the total work done by the gas is 13 kJ and the fraction of this work against the spring to compress it is 3 kJ.