---
title: "Piston with linear spring, air supply (15 kg)"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 1
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:piston-spring-air-15kg"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/piston-spring-air-15kg
site: BCT Engineering
---

# Piston with linear spring, air supply (15 kg)

Fundamental of Thermodynamics & Heat Transfer · Chapter 1: Introduction

- Asked in: 2071 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/piston-spring-air-15kg
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-1

## Question

A 15 kg piston in a cylinder with diameter of 0.15 m is loaded with a linear spring and the outside atmospheric pressure of 100 kPa, as shown in figure below. The spring exerts no force on the piston when it is at the lower position of the cylinder and for the state shown, the pressure is 300 kPa with volume of 0.02 m³. The valve is opened to let some air in, causing the piston to rise 5 cm. Find the new pressure. [Take g = 9.81 m/s²]
[Figure: vertical cylinder containing air, closed by a piston held down by a linear spring above it; an air supply line with a valve connects to the cylinder.]

## Short answer

Force balance P = P_atm + mg/A + kx/A. Using initial state, spring constant 2993 N/m. A 5 cm rise adds 8.47 kPa, giving new pressure 308.5 kPa.

## Answer

**Given:** m = 15 kg, D = 0.15 m, P_atm = 100 kPa, P₁ = 300 kPa, V₁ = 0.02 m³, rise Δx = 5 cm = 0.05 m, g = 9.81 m/s².

Force balance: P A = P_atm A + m g + k x, with spring force zero at the bottom (x = V/A).

**Step 1: Area**
A = (π/4)(0.15)² = 0.017671 m²

**Step 2: Piston weight pressure**
m g/A = (15 × 9.81)/0.017671 = 8327 Pa = 8.327 kPa

**Step 3: Spring pressure at state 1**
k x₁/A = 300 − 100 − 8.327 = 191.67 kPa

**Step 4: Piston height**
x₁ = V₁/A = 0.02/0.017671 = 1.1318 m

**Step 5: Spring constant**
k = (191.67 × 1000 × 0.017671)/1.1318 = 2993 N/m

**Step 6: Increase in pressure for 5 cm rise**
ΔP = k Δx / A = (2993 × 0.05)/0.017671 = 8468 Pa = 8.47 kPa

P₂ = 300 + 8.47 = **308.5 kPa**

**Answer: New pressure ≈ 308.5 kPa**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the spring force when the piston is at the bottom?
   A. Maximum
   B. Equal to the piston weight
   C. Equal to atmospheric force
   D. Zero, so F = kx with x = V/A
   Answer: D. The spring exerts no force at the bottom, so spring force is k times piston height x = V/A. (Key point: Force balance)

2. What is k x1/A at state 1 (300 kPa, P_atm = 100 kPa, m g/A = 8.327 kPa)?
   A. 191.67 kPa
   B. 200 kPa
   C. 308.3 kPa
   D. 108.3 kPa
   Answer: A. k x1/A = 300 - 100 - 8.327 = 191.67 kPa. (Key point: Spring pressure at state 1)

3. With x1 = 1.1318 m and A = 0.017671 m^2, k is about
   A. 299 N/m
   B. 29 930 N/m
   C. 2993 N/m
   D. 191.7 N/m
   Answer: C. k = 191.67 x 1000 x 0.017671 / 1.1318 = 2993 N/m. (Key point: Spring constant)

4. The new pressure after a 5 cm rise is about
   A. 291.5 kPa
   B. 308.5 kPa
   C. 350 kPa
   D. 500 kPa
   Answer: B. dP = k dx/A = 2993 x 0.05 / 0.017671 = 8.47 kPa, so P2 = 308.5 kPa. (Key point: New pressure)

## Close variants set in other exams

- [Piston with linear spring, air supply (5 kg)](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Apiston-spring-air-pressure-100mm) — Fundamental of Thermodynamics & Heat Transfer, chapter 1, asked 2×

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