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Ch 18: Thermal Properties of Matter
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 18, Problema 9

A large cylindrical tank contains 0.7500.750 m3 of nitrogen gas at 2727°C and 7.50×1037.50\(\times\)10^3 Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.4100.410 m3 and the temperature is increased to 157157°C?

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Start by identifying the initial and final states of the gas. The initial state has a volume \( V_1 = 0.750 \, \text{m}^3 \), temperature \( T_1 = 27^\circ \text{C} \), and pressure \( P_1 = 7.50 \times 10^3 \, \text{Pa} \). The final state has a volume \( V_2 = 0.410 \, \text{m}^3 \) and temperature \( T_2 = 157^\circ \text{C} \).
Convert the temperatures from Celsius to Kelvin, as the ideal gas law requires temperatures in Kelvin. Use the formula \( T(K) = T(^\circ C) + 273.15 \). Thus, \( T_1 = 27 + 273.15 = 300.15 \, \text{K} \) and \( T_2 = 157 + 273.15 = 430.15 \, \text{K} \).
Apply the combined gas law, which relates the pressure, volume, and temperature of a gas: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). This equation assumes the amount of gas remains constant.
Rearrange the combined gas law to solve for the final pressure \( P_2 \): \( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \).
Substitute the known values into the equation: \( P_2 = 7.50 \times 10^3 \, \text{Pa} \times \frac{0.750 \, \text{m}^3}{0.410 \, \text{m}^3} \times \frac{430.15 \, \text{K}}{300.15 \, \text{K}} \). Calculate \( P_2 \) using these values to find the final pressure.

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Ideal Gas Law

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