Question

Two massless, frictionless pistons are inside a horizontal tube opened at both ends. A 10-cm-long thread connects the pistons. The cross-sectional area of the tube is 20 cm2. The pressure and temperature of gas between the pistons and the outside air are the same and are equal to P = 1.0×105N/m2 and T = 23 ∘C.

A.) At what temperature will the thread break if it breaks when the tension reaches 33 N ?

Answer 1

Solution) Given Length L = 10 cm = 0.1 m

Area A = 20 cm^2 = 20×10^(-4) m^2

Pressure Po = 1×10^(5) N/m^2

Temperature To = 23°C = 23 + 273 K = 296 K

Tension (Force ) , F = 33 N

Temperature T = ?

We have ideal gas equation PV = nRT

T = (PV)/(nR)

So P and n values are to be calculated

P = Po + (F/A)

P = 1×10^(5) + (33/(20×10^(-4)))

P = 1.165 × 10^(5) N/m^2

We have

PoV = nRTo

n = (PoV)/(RTo)

V = AL (volume = area × length)

n = (1×10^(5)×20×10^(-4)×0.1)/(8.314×296)

n = 0.0081 moles

So PV = nRT

T = (PV)/(nR)

T = (1.165×10^(5)×20×10^(-4)×0.1)/(0.0081×8.314)

T = 345.98 K = 346 K

In °C temperature T = 346 - 273 = 73°C

T = 73°C