Edits to “Statement”, “Answer”
en/5.5.7.md
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| ### Statement | |||
| − | $5.5.7.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $5.5.7.$ A gas is in a vessel at a pressure of 2 MPa and a temperature of 27 °C. After being heated by 50 °C, only half of the gas (by mass) remains in the vessel. Determine the steady-state pressure. | |||
| ### Solution | |||
| Since the volume of the vessel remains constant, $V_1 = V_2 = V$. | |||
| Using the ideal gas law, the ratio of pressure and volume to mass and temperature is constant for a given gas: | |||
| $$\frac{p_1 V}{m_1 T_1} = \frac{p_2 V}{m_2 T_2}$$ | |||
| Expressing the final pressure $p_2$: | |||
| $$p_2 = p_1 \frac{T_2}{T_1} \frac{m_2}{m_1}$$ | |||
| We are given the initial pressure $p_1 = 2 \, \text{MPa} = 2 \cdot 10^6 \, \text{Pa}$. | |||
| The initial temperature is $T_1 = 27 \, ^\circ\text{C} = 300 \, \text{K}$. | |||
| The gas is heated by $50 \, ^\circ\text{C}$, so the final temperature is $T_2 = 300 + 50 = 350 \, \text{K}$. | |||
| The remaining mass is half of the initial mass, so $m_2 = \frac{m_1}{2}$. | |||
| Substituting these into the equation: | |||
| $$p_2 = \frac{p_1 T_2}{2 T_1}$$ | |||
| Calculating the value: | |||
| $$p_2 = \frac{2 \cdot 10^6 \cdot 350}{2 \cdot 300} \approx 1.166 \cdot 10^6 \, \text{Pa} = 1.166 \, \text{MPa}$$ | |||
| @@ -32,7 +26,3 @@Solution | |||
| #### Answer | |||
| $P = 1.166 \, \text{MPa}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $5.5.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $5.5.7.$ A gas is in a vessel at a pressure of 2 MPa and a temperature of 27 °C. After being heated by 50 °C, only half of the gas (by mass) remains in the vessel. Determine the steady-state pressure. | $5.5.7.$ A gas is in a vessel at a pressure of 2 MPa and a temperature of 27 °C. After being heated by 50 °C, only half of the gas (by mass) remains in the vessel. Determine the steady-state pressure. | ||
| ### Solution | ### Solution | ||
| Since the volume of the vessel remains constant, $V_1 = V_2 = V$. | Since the volume of the vessel remains constant, $V_1 = V_2 = V$. | ||
| Using the ideal gas law, the ratio of pressure and volume to mass and temperature is constant for a given gas: | Using the ideal gas law, the ratio of pressure and volume to mass and temperature is constant for a given gas: | ||
| $$\frac{p_1 V}{m_1 T_1} = \frac{p_2 V}{m_2 T_2}$$ | $$\frac{p_1 V}{m_1 T_1} = \frac{p_2 V}{m_2 T_2}$$ | ||
| Expressing the final pressure $p_2$: | Expressing the final pressure $p_2$: | ||
| $$p_2 = p_1 \frac{T_2}{T_1} \frac{m_2}{m_1}$$ | $$p_2 = p_1 \frac{T_2}{T_1} \frac{m_2}{m_1}$$ | ||
| We are given the initial pressure $p_1 = 2 \, \text{MPa} = 2 \cdot 10^6 \, \text{Pa}$. | We are given the initial pressure $p_1 = 2 \, \text{MPa} = 2 \cdot 10^6 \, \text{Pa}$. | ||
| The initial temperature is $T_1 = 27 \, ^\circ\text{C} = 300 \, \text{K}$. | The initial temperature is $T_1 = 27 \, ^\circ\text{C} = 300 \, \text{K}$. | ||
| The gas is heated by $50 \, ^\circ\text{C}$, so the final temperature is $T_2 = 300 + 50 = 350 \, \text{K}$. | The gas is heated by $50 \, ^\circ\text{C}$, so the final temperature is $T_2 = 300 + 50 = 350 \, \text{K}$. | ||
| The remaining mass is half of the initial mass, so $m_2 = \frac{m_1}{2}$. | The remaining mass is half of the initial mass, so $m_2 = \frac{m_1}{2}$. | ||
| Substituting these into the equation: | Substituting these into the equation: | ||
| $$p_2 = \frac{p_1 T_2}{2 T_1}$$ | $$p_2 = \frac{p_1 T_2}{2 T_1}$$ | ||
| Calculating the value: | Calculating the value: | ||
| $$p_2 = \frac{2 \cdot 10^6 \cdot 350}{2 \cdot 300} \approx 1.166 \cdot 10^6 \, \text{Pa} = 1.166 \, \text{MPa}$$ | $$p_2 = \frac{2 \cdot 10^6 \cdot 350}{2 \cdot 300} \approx 1.166 \cdot 10^6 \, \text{Pa} = 1.166 \, \text{MPa}$$ | ||
| @@ -32,7 +26,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $P = 1.166 \, \text{MPa}$ | $P = 1.166 \, \text{MPa}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||