91Ó°ÊÓ

FIGURE CP$21.70$shows two insulated compartments separated by a thin wall. The left side contains $0.060\mathrm{mol}$of helium at an initial temperature of $600\mathrm{K}$and the right side contains $0.030\mathrm{mol}$of helium at an initial temperature of $300\mathrm{K}$. The compartment on the right is attached to a vertical cylinder, above which the air pressure is $1.0\mathrm{atm}$. A $10-\mathrm{cm}$-diameter,$2.0\mathrm{kg}$ piston can slide without friction up and down the cylinder. Neither the cylinder diameter nor the volumes of the compartments are known.

a. What is the final temperature?

b. How much heat is transferred from the left side to the right side?

c. How high is the piston lifted due to this heat transfer?

d. What fraction of the heat is converted into work?

Step by step solution

01

Calculation for final temperature (part a)

a.

The heat is,

${\mathrm{Q}}_{12}=\frac{3}{2}{\mathrm{n}}_1\mathrm{R}\left({\mathrm{T}}_1-\mathrm{T}\right)$

${\mathrm{Q}}_{21}=\frac{5}{2}{\mathrm{n}}_2\mathrm{R}\left(\mathrm{T}-{\mathrm{T}}_2\right)$

Heat received by the system one is equal to the heat received by system .

two.

So,

$\frac{3}{2}{\mathrm{n}}_1\mathrm{R}\left({\mathrm{T}}_1-\mathrm{T}\right)=\frac{5}{2}{\mathrm{n}}_2\mathrm{R}\left(\mathrm{T}-{\mathrm{T}}_2\right).$

rearrange it,

$\mathrm{T}\left(5{\mathrm{n}}_2+3{\mathrm{n}}_1\right)=3{\mathrm{n}}_1{\mathrm{T}}_1+5{\mathrm{n}}_2{\mathrm{T}}_2$

Final temperature is ,

$\mathrm{T}=\frac{3{\mathrm{n}}_1{\mathrm{T}}_1+5{\mathrm{n}}_2{\mathrm{T}}_2}{5{\mathrm{n}}_2+3{\mathrm{n}}_1}$

Where

${\mathrm{n}}_1=0.060\mathrm{mol}$

${\mathrm{n}}_2=0.030\mathrm{mol}$

${\mathrm{T}}_1=600\mathrm{K}$

${\mathrm{T}}_2=300\mathrm{K}$

$\mathrm{T}=\frac{3×0.060\mathrm{mol}×600\mathrm{K}+5×0.030\mathrm{mol}×300\mathrm{K}}{5×0.030\mathrm{mol}+3×0.060\mathrm{mol}}$

$\mathrm{T}=464\mathrm{K}$

02

Explanation (part b)

b.

Where,

${\mathrm{Q}}_{12}=\frac{3}{2}{\mathrm{n}}_1\mathrm{R}\left({\mathrm{T}}_1-\mathrm{T}\right)$

${\mathrm{Q}}_{12}=\frac{3}{2}×0.060\mathrm{mol}×0.1\mathrm{m}\left(600\mathrm{K}-464\mathrm{K}\right)$

${\mathrm{Q}}_{12}=102\mathrm{J}$

03

Explanation (part c)

c.

Thermal energy is,

${\mathrm{Δ·¡}}_{\mathrm{th}2}=\frac{3}{2}{\mathrm{n}}_2\mathrm{R}\left(\mathrm{T}-{\mathrm{T}}_2\right)$

${\mathrm{Δ·¡}}_{\mathrm{th}2}=\frac{3}{2}×0.030\mathrm{mol}×0.1\mathrm{m}\left(464\mathrm{K}-300\mathrm{K}\right)$

$=61.2\mathrm{J}$

Work is,

$\mathrm{W}={\mathrm{Q}}_{12}-{\mathrm{Δ·¡}}_{\mathrm{th}2}$

$\mathrm{W}=102\mathrm{J}-61.2\mathrm{J}$

localid="1650297995208" $=40.8\mathrm{J}$

Distance is,

$\mathrm{h}=\frac{4\mathrm{W}}{{\mathrm{p}}_2{\mathrm{d}}^2\mathrm{Ï€}}$

$\mathrm{h}=\frac{4\mathrm{W}}{\left({\mathrm{p}}_{\mathrm{atm}}+\frac{4\mathrm{mg}}{{\mathrm{d}}^2\mathrm{Ï€}}\right){\mathrm{d}}^2\mathrm{Ï€}}$

$=\frac{4\mathrm{W}}{{\mathrm{p}}_{\mathrm{atm}}{\mathrm{d}}^2\mathrm{Ï€}+4\mathrm{mg}}$

role="math" localid="1650298160659" $=\frac{4×40.8\mathrm{J}}{1×{(0.1\mathrm{m})}^2\mathrm{π}+4×2\mathrm{Kg}}$

$\mathrm{h}=5.1\mathrm{cm}$

04

Calculation for efficiency (part d)

d.

The heat that is converted into work is,

$\mathrm{η}=\frac{\mathrm{W}}{{\mathrm{Q}}_{12}}$

$=\frac{40.8\mathrm{J}}{102\mathrm{J}}$

$=40\%$

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