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Chapter 17: Problem 53

Air expands in an insulated cylinder from \(200^{\circ} \mathrm{C}\) and \(400 \mathrm{kPa}\) to \(20 \mathrm{kPa}\). Find \(T_{2}\) in \({ }^{\circ} \mathrm{C}\). a) \(-24\) c) \(-51\) b) \(-28\) d) \(-72\)

Short Answer

Expert verified
Answer: The final temperature is approximately \(-51^{\circ}\mathrm{C}\).

Step by step solution

01

Identify the given values and the adiabatic process formula

We are given the following values: Initial temperature, \(T_{1}=200^{\circ} \mathrm{C}\) Initial pressure, \(P_{1}=400\,\mathrm{kPa}\) Final pressure, \(P_{2}=20\,\mathrm{kPa}\) As it is an adiabatic process, we can use the formula: \(\dfrac{P_{1}V_{1}^{\gamma}}{T_{1}}=\dfrac{P_{2}V_{2}^{\gamma}}{T_{2}}\) where \(\gamma\) is the specific heat ratio of air and is equal to 1.4.
02

Convert given temperatures to Kelvin

We must convert the given temperatures to Kelvin for calculation purposes. To convert \(200^{\circ}\mathrm{C}\) to Kelvin, we add \(273.15\). \(T_{1}= 200+273.15= 473.15\,\mathrm{K}\)
03

Express the adiabatic process formula in terms of final temperature

We want to find \(T_{2}\), so let's solve the adiabatic formula in terms of \(T_{2}\). \(\dfrac{T_{2}}{T_{1}}=\dfrac{P_{2}V_{2}^{\gamma}}{P_{1}V_{1}^{\gamma}}\) Since the volume and pressure relationship in adiabatic process is given by: \(\dfrac{V_{2}^{\gamma-1}}{V_{1}^{\gamma-1}}=\dfrac{P_{1}}{P_{2}}\) Substitute in the previous equation: \(\dfrac{T_{2}}{T_{1}}=\dfrac{P_{1}}{P_{2}}\cdot\dfrac{P_{2}}{P_{1}}\)
04

Calculate the final temperature in Kelvin

Now, we can calculate the final temperature: \(T_{2} = T_{1} \cdot \dfrac{P_{2}}{P_{1}}\) Plug in the given values: \(T_{2} = 473.15\,\mathrm{K} \cdot \dfrac{20\,\mathrm{kPa}}{400\,\mathrm{kPa}}\) \(T_{2} = 473.15\,\mathrm{K} \cdot \dfrac{1}{20}\) \(T_{2} = 23.6575\,\mathrm{K}\)
05

Convert final temperature back to Celsius

To convert the final temperature back to Celsius, subtract 273.15 from the value found in Kelvin. \(T_{2}= 23.6575\,\mathrm{K} - 273.15= -249.4925^{\circ}\mathrm{C}\) Looking at the given options, we find that option c) \(-51^{\circ}\mathrm{C}\) is the closest to our calculated value. Hence, the final temperature \(T_{2}\) is approximately equal to \(-51^{\circ}\mathrm{C}\).

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