Giữ nhiệt độ ở phần 2 không đổi, nung nóng phần 1 đến nhiệt độ 600 K thì pit-tông cách đáy dưới khoảng là h. Tìm tỉ số h/L (làm tròn kết quả đến chữ số hàng phần tră

Đáp án: 0,45

+ $\frac{{\text{p}}_1^{\text{'}}}{{\text{p}}_1}=\frac{{\text{V}}_1}{{\text{ V}}_1^{\text{'}}}.\frac{{\text{T}}_1^{\text{'}}}{{\text{T}}_1}=\frac{0,4\text{LS}}{\left(\text{ L}-\text{h}\right).\text{S}}.2\Rightarrow {\text{p}}_1^{\text{'}}=\frac{1440\text{ L}}{\text{ L}-\text{h}}$

+ ${\text{T}}_2={\text{T}}_2^{\text{'}}\Rightarrow \frac{{\text{p}}_2^{\text{'}}}{{\text{p}}_2}=\frac{{\text{V}}_2}{{\text{ V}}_2^{\text{'}}}\Rightarrow \frac{{\text{p}}_2^{\text{'}}}{{\text{p}}_2}=\frac{0,6\text{ L}.\text{ S}}{\text{ h}.\text{S}}\Rightarrow {\text{p}}_2^{\text{'}}=\frac{1440\text{ L}}{\text{ h}}$

${\text{p}}_2^{\text{'}}-{\text{p}}_1^{\text{'}}=600\Rightarrow \frac{1440}{\frac{\text{ h}}{\text{ L}}}-\frac{1440}{1-\frac{\text{ h}}{\text{ L}}}=600$

\( \Rightarrow \frac{{\rm{h}}}{{\rm{L}}} = 0,45\)

Hoặc:

+ ${\text{p}}_2^{\text{'}}-{\text{p}}_1^{\text{'}}=\frac{\text{mg}}{\text{S}}=600\left(\text{Pa}\right)$

+ $\frac{{\text{p}}_2^{\text{'}}{\text{V}}_2^{\text{'}}}{{\text{T}}_2^{\text{'}}}=\frac{{\text{p}}_2{\text{V}}_2}{{\text{ T}}_2}\Rightarrow \frac{{\text{p}}_2^{\text{'}}.\left(0,6\text{L}-\text{k}\right)\text{S}}{27+273}=\frac{2400.0,6\text{L}.\text{S}}{27+273}$

$\Rightarrow {\text{p}}_2^{\text{'}}=\frac{1440.\text{ L}}{0,6\text{L}-\text{k}}$

+ $\frac{{\text{p}}_1^{\text{'}}.{\text{V}}_1^{\text{'}}}{{\text{T}}_1^{\text{'}}}=\frac{{\text{p}}_1{\text{V}}_1}{{\text{ T}}_1}\Rightarrow \frac{{\text{p}}_1^{\text{'}}.\left(0,4\text{L}+\text{k}\right).\text{S}}{600}=\frac{1800.0,4\text{L}.\text{S}}{27+273}$

$\Rightarrow {\text{p}}_1^{\text{'}}=\frac{1440\text{ L}}{0,4\text{L}+\text{k}}$

\( \Rightarrow \frac{{1440{\rm{L}}}}{{0,6{\rm{L}} - {\rm{k}}}} - \frac{{1440{\rm{L}}}}{{0,4{\rm{L}} + {\rm{k}}}} = 600\)

\( \Rightarrow \frac{{1440}}{{0,6 - \frac{{\rm{k}}}{{\rm{L}}}}} - \frac{{1440}}{{0,4 + \frac{{\rm{k}}}{{\rm{L}}}}} = 600\)

\( \Rightarrow \frac{{\rm{k}}}{{\rm{L}}} \approx 0,15 \Rightarrow \frac{{{\rm{\;h}}}}{{{\rm{\;L}}}} \approx 0,45\)

\(\left( {{\rm{h}} = 0,6 - {\rm{k}};\frac{{\rm{h}}}{{\rm{L}}} = 0,6 - \frac{{\rm{k}}}{2}} \right)\)