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**В расположенном горизонтально теплоизолированном жёстком цилиндре может перемешаться поршень, по одну сторону от которого находится $\nu$ молей идеального газа споказателем адиабаты $\gamma$, a по другую --- вакуум. Между поршнем и дном цилиндра помещена пружина. В начальный момент поршень закреплён, а пружина не деформирована. Затем поршень освобождают. После установления равновесия объём газа увеличился в $n$ раз. Определить изменение энтропии газа. При расчёте пренебречь трением, а также теплоёмкостями цилиндра, поршня и пружины. Считать, что к деформациям пружины применим закон Гука.**
---
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p
$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow
$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_pE_p = \frac{k(nl_0)^2}{2}
$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2}
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2}\\
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = - \frac{k\big((n-1)l_0\big)^2}{2}
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2}
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2} \Rightarrow \\
\Rightarrow \frac{\Delta T}{\gamma -1}& = - \frac{T_1(n-1)}{2n}
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2} \Rightarrow \\
\Rightarrow \frac{\Delta T}{\gamma -1}& = - \frac{T_1}{2n} \Rightarrow T_0 = T_1 \bigg(1 + \frac{(\gamma -1)(n-1)}{2n}\bigg)
\end{align}
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2} \Rightarrow \\
\Rightarrow \frac{\Delta T}{\gamma -1}& = - \frac{T_1}{2n} \Rightarrow T_0 = T_1 \bigg(1 + \frac{(\gamma -1)(n-1)}{2n}\bigg)
\end{align}
$$\Delta S = \nu \bigg( R \ln{n} + C_V \ln{\frac{2n}{\gamma n + n - \gamma + 1}}\bigg)$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2} \Rightarrow \\
\Rightarrow \frac{\Delta T}{\gamma -1}& = - \frac{T_1}{2n} \Rightarrow T_0 = T_1 \bigg(1 + \frac{(\gamma -1)(n-1)}{2n}\bigg)
\end{align}
$$\Delta S = \nu \bigg( R \ln{n} + C_V \ln{\frac{2n}{\gamma n + n - \gamma +1}}\bigg) = \nu R \bigg( \ln{n} + \frac{1}{\gamma -1} \cdot \ln{\frac{2n}{\gamma n + n - \gamma + 1}}\bigg)$$
----
$$\Delta S = \nu \bigg(R\ln{\frac{V_1}{V_0}} + C_V\ln{\frac{T_1}{T_0}}\bigg)$$
$$E_0 = E_1 + E_p \Rightarrow \Delta E = - E_p;
$$
\begin{align}
E_p&= \frac{k\big((n-1)l_0\big)^2}{2};\\
p_1 S& = k(n-1)l_0;\\
p_1 V_1& = \nu RT_1; V_1 = nl_0S\\
\Delta E& = \frac{\nu R \Delta T}{\gamma -1}
\end{align}
\begin{align}
\frac{\nu R T_1 S}{n l_0 S}& = k(n-1)l_0 \Rightarrow k = \frac{\nu R T_1 }{n(n-1)l_0^2};\\
\frac{\nu R \Delta T}{\gamma -1}& = \frac{k\big((n-1)l_0\big)^2}{2} \Leftrightarrow \frac{\nu R \Delta T}{\gamma -1} = - \frac{\nu R T_1 }{n(n-1)l_0^2}\cdot \frac{\big((n-1)l_0\big)^2}{2} \Rightarrow \\
\Rightarrow \frac{\Delta T}{\gamma -1}& = - \frac{T_1}{2n} \Rightarrow T_0 = T_1 \bigg(1 + \frac{(\gamma -1)(n-1)}{2n}\bigg)
\end{align}
$$\Delta S = \nu \bigg( R \ln{n} + C_V \ln{\frac{2n}{\gamma n + n - \gamma +1}}\bigg) = \boxed{\nu R \bigg( \ln{n} + \frac{1}{\gamma -1} \cdot \ln{\frac{2n}{\gamma n + n - \gamma + 1}}\bigg)}$$
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