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Chapter 8: Problem 1181

Air is filled in a motor tube at \(27^{\circ} \mathrm{C}\) and at a Pressure of 8 atmosphere. The tube suddenly bursts. Then what is the temperature of air. given \(\gamma\) of air \(=1.5\) (A) \(150 \mathrm{~K}\) (B) \(150^{\circ} \mathrm{C}\) (C) \(75 \mathrm{~K}\) (D) \(27.5^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature of the air after the motor tube bursts is approximately \(150 \mathrm{K}\) (Option A).

Step by step solution

01

Convert the given values to appropriate units

We are given the initial temperature as \(27^{\circ}\mathrm{C}\). To convert it into Kelvin, we use the relation, \(T_1(kelvin) = T_1(celsius) + 273\). Also, we are given that the initial pressure is 8 atmospheres. The SI unit of pressure is Pascal. To convert it to Pascal, we can use the following relation: \[1 \,atm = 101325 \,Pa\]
02

Calculate the initial temperature in Kelvin

Convert the given temperature (in Celsius) to Kelvin: \(T_1 = 27^{\circ}\mathrm{C} + 273 = 300 \,K\)
03

Calculate the initial pressure in Pascals

Convert the given pressure (in atmospheres) to Pascal: \(P_1 = 8\,atm \times 101325\,Pa/atm = 810600\,Pa\)
04

Set the final pressure

The motor tube bursts, and air pressure outside is considered as 1 atmosphere. \(P_2 = 1 \,atm = 101325\,Pa\)
05

Calculate the final temperature using the adiabatic process formula

Use the given adiabatic equation to find the final temperature: \(T_2 = T_1 \times \left(\frac{P_1}{P_2}\right)^{(\gamma - 1) / \gamma}\) \(T_2 = 300\,K \times \left(\frac{810600\,Pa}{101325\,Pa}\right)^{(1.5 - 1) / 1.5}\)
06

Calculate the final temperature and convert it back to Celsius

After solving the above equation, we get: \(T_2 = 150\,K\) Now, convert the final temperature back to Celsius: \(T_2 = 150\,K - 273 = -123^{\circ}\mathrm{C}\) The final temperature after the motor tube bursts is \(-123^{\circ}\mathrm{C}\), which is not among the given options. However, if we closely inspect the options, we notice that the temperature of \(75\,K\) and \(150\,K\) are quite close. So, there must be some approximation in the given options. Therefore, option (A) is the closest and can be considered as the correct answer.

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Most popular questions from this chapter

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

A difference of temperature of \(25^{\circ} \mathrm{Cis}\) equivalent to a difference of (A) \(72^{\circ} \mathrm{F}\) (B) \(45^{\circ} \mathrm{F}\) (C) \(32^{\circ} \mathrm{F}\) (D) \(25^{\circ} \mathrm{F}\)

Starting with the same initial Conditions, an ideal gas expands from Volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) in three different ways. The Work done by the gas is \(\mathrm{W}_{1}\) if the process is purely isothermal, \(\mathrm{W}_{2}\) if purely isobaric and \(\mathrm{W}_{3}\) if purely adiabatic Then (A) \(\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}\) (B) \(\mathrm{W}_{2}>\mathrm{W}_{3}>\mathrm{W}_{1}\) (C) \(\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}\) (D) \(\mathrm{W}_{1}>\mathrm{W}_{3}>\mathrm{W}_{2}\)

If a heat engine absorbs \(50 \mathrm{KJ}\) heat from a heat source and has efficiency of \(40 \%\), then the heat released by it in heat sink is (A) \(40 \mathrm{KJ}\) (B) \(30 \mathrm{KJ}\) (C) \(20 \mathrm{~J}\) (D) \(20 \mathrm{KJ}\)

A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registered \(140^{\circ} .\) What is the fall in thermometers (A) \(80^{\circ}\) (B) \(60^{\circ}\) (C) \(40^{\circ}\) (D) \(30^{\circ}\)
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