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- 0. Math Review31m
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22. The First Law of Thermodynamics
First Law of Thermodynamics
18:02 minutes
Problem 19.61a
Textbook Question
Textbook QuestionTwo cylinders each contain 0.10 mol of a diatomic gas at 300 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.
a. What are the final temperature and volume of each?
Verified step by step guidance
1
Identify the initial conditions for both cylinders: initial number of moles (n = 0.10 mol), initial temperature (T_i = 300 K), and initial pressure (P_i = 3.0 atm).
For Cylinder A (isothermal process): Use the ideal gas law, PV = nRT, to find the initial volume. Since the temperature remains constant in an isothermal process, apply the formula P_iV_i = P_fV_f to find the final volume V_f, where P_f is the final pressure (1.0 atm).
For Cylinder B (adiabatic process): Use the adiabatic condition, where PV^\gamma = constant (\gamma is the heat capacity ratio, which is approximately 1.4 for diatomic gases). Use the initial conditions to find the constant, and then rearrange to find the final volume V_f using the final pressure P_f = 1.0 atm.
To find the final temperature of Cylinder B, use the adiabatic condition T_iV_i^{\gamma-1} = T_fV_f^{\gamma-1}. Solve for T_f using the known values of T_i, V_i, V_f, and \gamma.
Summarize the final conditions: For Cylinder A, the final temperature T_f is the same as the initial (300 K) and the final volume is calculated from step 2. For Cylinder B, the final temperature is calculated from step 4 and the final volume from step 3.
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Textbook Question
Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 * 10^6 J/kg and the boiling point is 120°C. At this pressure, 1.00 kg of water has a volume of 1.00 * 10^-3 m^3 , and 1.00 kg of steam has a volume of 0.824 m^3. (b) Compute the increase in internal energy of the water.
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