Question

A 10 cm -diameter cylinder contains argon gas at 10 atm pressure and a temperature of 60 ∘C . A piston can slide in and out of the cylinder. The cylinder's initial length is 19 cm . 2600 J of heat are transferred to the gas, causing the gas to expand at constant pressure. 1-What is the final temperature of the cylinder? 2-What is the final length of the cylinder?

Answer #1

1. Final temp of the cylinder in an isobaric (const. pressure) heat transfer process is given by :-

Q = nCpT = 1*0.520kJ/(kg K))*(Tf - Ti)

Assuming it to be an ideal gas,

PV = nRT

1013250 * (3.14*0.05*0.05*0.19) = n * 8.3145 * 333

n = 0.54

2.6kJ = 0.54*0.520kJ/(kg K))*(Tf - 333K)

(Tf - 333K) = 2.6 / 0.28 = 9.28

Tf = 342.28K

2. Final length of cylinder will be given by :-

PV=nRT

Given, P =constant. So, V/T = constant

V1 / T1 = V2 / T2

(3.14*0.05*0.05*0.19) / 333 = V2 / 342.28

V2 = 0.00153 m3

0.00153 = 3.14*0.05*0.05*L

L = 19.52 cm

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