Any help would be appreciated. How to relate length change to pressure change? T

Question

Any help would be appreciated. How to relate length change to pressure change? The space above the mercury column in a thermometer ordinarily is evacuated, but due to faulty manufacture, a particular thermometer has a pressure of 2 mmHg of air in this space when the hole thermometer is immersed in a bath at 0 degree C Calculate the pressure of the air when the whole thermometer is immersed in a bath at 100 degree C. As the figure below shows, at 0 degree C. the length of air space is 10 cm. and at 100 degree C. it is 2 cm.

Explanation / Answer

let initial volume be V
then final volume = V/5 (since the length of air becomes 2cm as compared to 10cm earlier)

Ti = 0+ 273 = 273K
Tf = 100+273 = 373 K

number of moles is constant inside
PV/T = nR
=>PV/T is constant

now initial pressure Pi = 2mm hg
so, 2*V/273 = (Pf*V/5)/373 (Pf is the final pressure of air)
=> Pf = 13.66 mm hg now let the change in length be x initial length be l and final length be l+x let initial volume be V now volume is proportional to length since area of cross section is constant here so, final volume V' = V(l+x)/l PV/T is constant so, PiV/Ti = PfV(l+x)/lTf =>PiTf/Ti = Pf(1+x/l) =>PiTf/Ti(1+x/l) = Pf now pressure change p = Pf-Pi = Pi(Tf/Ti(1+x/l) -1 ) let initial volume be V
then final volume = V/5 (since the length of air becomes 2cm as compared to 10cm earlier)

Ti = 0+ 273 = 273K
Tf = 100+273 = 373 K

number of moles is constant inside
PV/T = nR
=>PV/T is constant

now initial pressure Pi = 2mm hg
so, 2*V/273 = (Pf*V/5)/373 (Pf is the final pressure of air)
=> Pf = 13.66 mm hg then final volume = V/5 (since the length of air becomes 2cm as compared to 10cm earlier)

Ti = 0+ 273 = 273K
Tf = 100+273 = 373 K

number of moles is constant inside
PV/T = nR
=>PV/T is constant

now initial pressure Pi = 2mm hg
so, 2*V/273 = (Pf*V/5)/373 (Pf is the final pressure of air)
=> Pf = 13.66 mm hg now initial pressure Pi = 2mm hg => Pf = 13.66 mm hg now let the change in length be x initial length be l and final length be l+x let initial volume be V now volume is proportional to length since area of cross section is constant here so, final volume V' = V(l+x)/l PV/T is constant so, PiV/Ti = PfV(l+x)/lTf =>PiTf/Ti = Pf(1+x/l) =>PiTf/Ti(1+x/l) = Pf

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