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Contests/Gnit Sunday Open 002 (GSO002)/Problem 6 First Law of Thermodynamics in a Bicycle Pump (Compression Before Valve Opens)
Problem 6

First Law of Thermodynamics in a Bicycle Pump (Compression Before Valve Opens)

Finished
200 ptsLv.3 ElementaryThermodynamics
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

Problem Statement

A cyclist uses a cylinder-type pump to inflate a bicycle tire. Consider the compression process while the inside of the cylinder is sealed (closed system) — from when the piston begins to be pushed in until just before the tire valve opens and air begins to enter the tire.

Treat the air inside the cylinder as a single thermodynamic system, modeled as an ideal gas with constant-volume molar heat capacity CV=52RC_V = \frac{5}{2}RCV​=25​R (RRR is the gas constant).

In the initial state, the air inside the cylinder has pressure P1P_1P1​ and volume V1V_1V1​. After the cyclist pushes the piston in, the air is compressed, and just before the valve opens, the pressure is P2P_2P2​ and the volume is V2V_2V2​.

Let WinW_{in}Win​ be the work done on the gas by the cyclist (external agent) during this compression. Due to the temperature rise caused by compression, the air releases a heat QoutQ_{out}Qout​ to the outside through the cylinder wall.

Find the heat Qout[J]Q_{out} [\mathrm{J}]Qout​[J] released by the air to the outside during this process.

Sign conventions:

  • P1,V1,P2,V2,Win,QoutP_1, V_1, P_2, V_2, W_{in}, Q_{out}P1​,V1​,P2​,V2​,Win​,Qout​ are all positive.
  • With QQQ = heat absorbed by gas from outside, WWW = work done by gas on outside, ΔU\Delta UΔU = change in internal energy: the first law is Q=ΔU+WQ = \Delta U + WQ=ΔU+W.

Constraints

  • Initial pressure: P1=1.0×105 PaP_1 = 1.0 \times 10^5\ \mathrm{Pa}P1​=1.0×105 Pa
  • Initial volume: V1=2.5×10−3 m3V_1 = 2.5 \times 10^{-3}\ \mathrm{m}^3V1​=2.5×10−3 m3
  • Final pressure (just before valve opens): P2=3.0×105 PaP_2 = 3.0 \times 10^5\ \mathrm{Pa}P2​=3.0×105 Pa
  • Final volume (just before valve opens): V2=1.48×10−3 m3V_2 = 1.48 \times 10^{-3}\ \mathrm{m}^3V2​=1.48×10−3 m3
  • Work done on gas by external agent: Win=562 JW_{in} = 562\ \mathrm{J}Win​=562 J

Input Format

Give the value of Qout[J]Q_{out} [\mathrm{J}]Qout​[J] as a positive integer.

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