GSO002 Problem 6
Problem Statement
First Law of Thermodynamics in a Bicycle Pump (Compression Before Valve Opens)
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=25R (R is the gas constant).
In the initial state, the air inside the cylinder has pressure P1 and volume V1. After the cyclist pushes the piston in, the air is compressed, and just before the valve opens, the pressure is P2 and the volume is V2.
Let 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 Qout to the outside through the cylinder wall.
Find the heat Qout[J] released by the air to the outside during this process.
Sign conventions:
- P1,V1,P2,V2,Win,Qout are all positive.
- With Q = heat absorbed by gas from outside, W = work done by gas on outside, ΔU = change in internal energy: the first law is Q=ΔU+W.
Constraints
- Initial pressure: P1=1.0×105 Pa
- Initial volume: V1=2.5×10−3 m3
- Final pressure (just before valve opens): P2=3.0×105 Pa
- Final volume (just before valve opens): V2=1.48×10−3 m3
- Work done on gas by external agent: Win=562 J
Input Format
Give the value of Qout[J] as a positive integer.
Solution
1. Internal Energy Formula
The internal energy U of an ideal gas with molar heat capacity CV, temperature T, and amount n is:
U=nCVTSince the cylinder is sealed, n is constant. With CV=25R:
U=25nRTUsing the ideal gas law PV=nRT:
U=25PV2. Computing the Change in Internal Energy ΔU
ΔU=U2−U1=25(P2V2−P1V1)Substituting:
P1V1=(1.0×105)×(2.5×10−3)=250 J P2V2=(3.0×105)×(1.48×10−3)=444 J ΔU=25(444−250)=25(194)=5×97=485 J3. Applying the First Law to Find Qout
The gas releases heat Qout: Q=−Qout. The gas receives work from outside: W=−Win.
Substituting into the first law Q=ΔU+W:
−Qout=ΔU+(−Win) Qout=Win−ΔUThis means the portion of the external work that is not stored as internal energy is released as heat.
Qout=562−485=77 JAnswer: 77