New solution
en/5.9.4.md
+37 −0
| @@ -0,0 +1,37 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $5.9.4.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $5.9.4.$ Find the entropy increment of hydrogen when it expands from volume $V$ to $2V$: a) in a vacuum; b) in an isothermal process. The mass of the gas is $m$. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | In the general case, the change in entropy of an ideal gas when transitioning from state $1$ to state $2$ is described by the formula: | ||
| + | $$\Delta S = \frac{m}{M} C_V \ln\left(\frac{T_2}{T_1}\right) + \frac{m}{M} R \ln\left(\frac{V_2}{V_1}\right)$$ | ||
| + | where $\frac{m}{M}$ is the amount of substance (moles), $C_V$ is the molar heat capacity at constant volume, and $R$ is the universal gas constant. | ||
| + | |||
| + | Let us consider both cases: | ||
| + | |||
| + | <b>b) In an isothermal process</b> | ||
| + | According to the given conditions, the process is isothermal, hence the temperature is constant ($T_1 = T_2$). The first term in the entropy equation (which depends on temperature) becomes zero. | ||
| + | The change in entropy is determined solely by the change in volume from $V_1 = V$ to $V_2 = 2V$: | ||
| + | $$\Delta S_b = \frac{m}{M} R \ln\left(\frac{2V}{V}\right) = \frac{m}{M} R \ln 2$$ | ||
| + | |||
| + | <b>a) Upon expansion in a vacuum</b> | ||
| + | The expansion of a gas into a vacuum is an adiabatic process with no work done ($Q = 0$, $A = 0$). According to the first law of thermodynamics, the change in internal energy is $\Delta U = Q - A = 0$. | ||
| + | Since the internal energy of an ideal gas depends only on its temperature, a constant internal energy implies that the gas temperature remains unchanged ($T_1 = T_2$). | ||
| + | |||
| + | Thus, the initial and final states of the gas upon expansion in a vacuum are exactly the same as in the isothermal expansion. Since entropy is a <b>state function</b>, its change does not depend on the path the system took to get from the initial state to the final one. | ||
| + | Therefore, the entropy increment in a vacuum will be exactly the same: | ||
| + | $$\Delta S_a = \Delta S_b = \frac{m}{M} R \ln 2$$ | ||
| + | |||
| + | #### Answer | ||
| + | a, b: $\Delta S = \frac{m}{M} R \ln 2$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,37 @@ | |||
| ### Statement | |||
| $5.9.4.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $5.9.4.$ Find the entropy increment of hydrogen when it expands from volume $V$ to $2V$: a) in a vacuum; b) in an isothermal process. The mass of the gas is $m$. | |||
| ### Solution | |||
| In the general case, the change in entropy of an ideal gas when transitioning from state $1$ to state $2$ is described by the formula: | |||
| $$\Delta S = \frac{m}{M} C_V \ln\left(\frac{T_2}{T_1}\right) + \frac{m}{M} R \ln\left(\frac{V_2}{V_1}\right)$$ | |||
| where $\frac{m}{M}$ is the amount of substance (moles), $C_V$ is the molar heat capacity at constant volume, and $R$ is the universal gas constant. | |||
| Let us consider both cases: | |||
| <b>b) In an isothermal process</b> | |||
| According to the given conditions, the process is isothermal, hence the temperature is constant ($T_1 = T_2$). The first term in the entropy equation (which depends on temperature) becomes zero. | |||
| The change in entropy is determined solely by the change in volume from $V_1 = V$ to $V_2 = 2V$: | |||
| $$\Delta S_b = \frac{m}{M} R \ln\left(\frac{2V}{V}\right) = \frac{m}{M} R \ln 2$$ | |||
| <b>a) Upon expansion in a vacuum</b> | |||
| The expansion of a gas into a vacuum is an adiabatic process with no work done ($Q = 0$, $A = 0$). According to the first law of thermodynamics, the change in internal energy is $\Delta U = Q - A = 0$. | |||
| Since the internal energy of an ideal gas depends only on its temperature, a constant internal energy implies that the gas temperature remains unchanged ($T_1 = T_2$). | |||
| Thus, the initial and final states of the gas upon expansion in a vacuum are exactly the same as in the isothermal expansion. Since entropy is a <b>state function</b>, its change does not depend on the path the system took to get from the initial state to the final one. | |||
| Therefore, the entropy increment in a vacuum will be exactly the same: | |||
| $$\Delta S_a = \Delta S_b = \frac{m}{M} R \ln 2$$ | |||
| #### Answer | |||
| a, b: $\Delta S = \frac{m}{M} R \ln 2$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||