- #1

maverick280857

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Hello everyone

My query is regarding the following problem:

A barometer reads 75.0 cm on a steel scale. The room temperature is 30 degrees C. The scale is correctly graduated for 0 degrees C. The coefficient of linear expansion of steel is [itex]\alpha = 1.2 * 10^{-5}[/itex] per degree C and the coefficient of volume expansion of mercury is [itex]\gamma = 1.8 * 10^{-4}[/itex] per degree C. Find the correct atmospheric pressure.

My solution does not match the book's solution, which goes as follows:

the length of the 75 cm steel wire at 0 degrees C will become [itex]l_{\theta}[/itex] where,

[tex]l_{\theta} = (75cm)(1 + \alpha(30))[/tex]

The length of the mercury column at 30 degrees C is [itex]l_{\theta}[/itex]. Suppose the length of the mercury column if it were at 0 degrees C is [itex]l_{0}[/itex]. Then,

[tex]l_{\theta} = l_{0}(1 +\frac{\gamma}{3}(30))[/tex]

So

[tex]l_{0}(1 + \frac{\gamma}{3}(30)) = (75 cm)(1 + \alpha(30))[/tex]

or [tex]l_{0} = 75cm \frac{1 + \alpha(30)}{1 + \frac{\gamma}{3}(30)}[/tex] = 74.89 cm (after the binomial approximation)

Suppose the volume of mercury at the higher temperature is [itex]V_{Hg}[/itex], the cross-sectional area of the mercury column (equal to the cross-sectional area of the steel tube) is [itex]A_{v}[/itex] and the height of the mercury column is [itex]h_{Hg}[/itex].

[tex]V_{Hg} = V_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

[tex]A_{v} = A_{v,0}(1 + \beta_{v}\Delta T)[/tex]

where [itex]\gamma_{Hg}[/itex] and [itex]\beta_{v}[/itex] are the coefficients of volume expansion of mercury and of area expansion of the vessel (steel tube).

Also

[tex]V_{Hg} = A_{v}h_{Hg}[/tex]

and

[tex]V_{Hg,0} = A_{v,0}h_{Hg,0}[/tex]

[tex]A_{v}h_{Hg} = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

so

[tex]A_{v,0}h_{Hg}(1 + \beta_{v}\Delta T) = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

so

[tex]h_{Hg} = h_{Hg,0}\frac{(1 + \gamma_{Hg}\Delta T)}{(1 + \beta_{v}\Delta T)}[/tex]

which after the binomial approximation yields

[tex]h_{Hg} = h_{Hg,0}(1 + \gamma_{Hg}\Delta T)(1 - \beta_{v}\Delta T)[/tex]

If we neglect the term second order in [itex](\Delta T)^2[/itex], then

[tex]h_{Hg} = h_{Hg,0}(1 + (\gamma_{Hg}-\beta_{v})\Delta T)[/tex]

But [tex]\beta_{v} = \frac{\alpha_{v}}{2}[/tex]. Instead of the factor 1/3 I get 1/2. Where I am I going wrong?

Thanks and cheers

Vivek

My query is regarding the following problem:

A barometer reads 75.0 cm on a steel scale. The room temperature is 30 degrees C. The scale is correctly graduated for 0 degrees C. The coefficient of linear expansion of steel is [itex]\alpha = 1.2 * 10^{-5}[/itex] per degree C and the coefficient of volume expansion of mercury is [itex]\gamma = 1.8 * 10^{-4}[/itex] per degree C. Find the correct atmospheric pressure.

My solution does not match the book's solution, which goes as follows:

the length of the 75 cm steel wire at 0 degrees C will become [itex]l_{\theta}[/itex] where,

[tex]l_{\theta} = (75cm)(1 + \alpha(30))[/tex]

The length of the mercury column at 30 degrees C is [itex]l_{\theta}[/itex]. Suppose the length of the mercury column if it were at 0 degrees C is [itex]l_{0}[/itex]. Then,

[tex]l_{\theta} = l_{0}(1 +\frac{\gamma}{3}(30))[/tex]

So

[tex]l_{0}(1 + \frac{\gamma}{3}(30)) = (75 cm)(1 + \alpha(30))[/tex]

or [tex]l_{0} = 75cm \frac{1 + \alpha(30)}{1 + \frac{\gamma}{3}(30)}[/tex] = 74.89 cm (after the binomial approximation)

**My solution**Suppose the volume of mercury at the higher temperature is [itex]V_{Hg}[/itex], the cross-sectional area of the mercury column (equal to the cross-sectional area of the steel tube) is [itex]A_{v}[/itex] and the height of the mercury column is [itex]h_{Hg}[/itex].

[tex]V_{Hg} = V_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

[tex]A_{v} = A_{v,0}(1 + \beta_{v}\Delta T)[/tex]

where [itex]\gamma_{Hg}[/itex] and [itex]\beta_{v}[/itex] are the coefficients of volume expansion of mercury and of area expansion of the vessel (steel tube).

Also

[tex]V_{Hg} = A_{v}h_{Hg}[/tex]

and

[tex]V_{Hg,0} = A_{v,0}h_{Hg,0}[/tex]

[tex]A_{v}h_{Hg} = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

so

[tex]A_{v,0}h_{Hg}(1 + \beta_{v}\Delta T) = A_{v,0}h_{Hg,0}(1 + \gamma_{Hg}\Delta T)[/tex]

so

[tex]h_{Hg} = h_{Hg,0}\frac{(1 + \gamma_{Hg}\Delta T)}{(1 + \beta_{v}\Delta T)}[/tex]

which after the binomial approximation yields

[tex]h_{Hg} = h_{Hg,0}(1 + \gamma_{Hg}\Delta T)(1 - \beta_{v}\Delta T)[/tex]

If we neglect the term second order in [itex](\Delta T)^2[/itex], then

[tex]h_{Hg} = h_{Hg,0}(1 + (\gamma_{Hg}-\beta_{v})\Delta T)[/tex]

But [tex]\beta_{v} = \frac{\alpha_{v}}{2}[/tex]. Instead of the factor 1/3 I get 1/2. Where I am I going wrong?

Thanks and cheers

Vivek

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