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Chapter 8: Problem 1163

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) \(405 \mathrm{~K}\)

Short Answer

Expert verified
The rise in temperature is \(375\,K\), which corresponds to the choice (C).

Step by step solution

01

Write down the given variables

: Initial temperature, \(T_1\) = 27°C = 273 + 27 = 300 K Initial volume, \(V_1\) Final volume, \(V_2\) = 8/27 \(V_1\) Adiabatic Index, \(v\) = 5/3
02

Write down the equation for the adiabatic process

: We will use the equation for an adiabatic process of an ideal gas: \(\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{v-1}\)
03

Substitute given values into the equation

: We now substitute the given values into the equation: \(\frac{300}{T_2} = \left(\frac{8V_1 / 27V_1}{V_1}\right)^{5/3 - 1}\)
04

Solve for \(T_2\)

: Simplify the equation and solve for \(T_2\): \(\frac{300}{T_2} = \left(\frac{8}{27}\right)^{2/3}\) \(T_2 = 300 \times \left(\frac{27}{8}\right)^{2/3}\) \(T_2 = 300(1.5)^2\) Therefore, \(T_2 = 675\,K\)
05

Calculate the rise in temperature

: Now, we just need to find the difference between \(T_2\) and \(T_1\): Temperature rise = \(T_2 - T_1\) Temperature rise = \(675\,K - 300\,K\) Temperature rise = \(375\,K\) The rise in temperature is 375 K, which corresponds to the choice (C).

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Most popular questions from this chapter

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

In a container of negligible heat capacity, 200 g ice at $0^{\circ} \mathrm{C}\( and \)100 \mathrm{~g}\( steam at \)100^{\circ} \mathrm{C}$ are added to \(200 \mathrm{~g}\) of water that has temperature \(55^{\circ} \mathrm{C}\). Assume no heat is lost to the surroundings and the pressure in the container is constant \(1 \mathrm{~atm}\). At the final temperature, mass of the total water present in the system is (A) \(493.6 \mathrm{~g}\) (B) \(483.3 \mathrm{~g}\) (C) \(472.6 \mathrm{~g}\) (D) \(500 \mathrm{~g}\)

The isothermal Bulk modulus of an ideal gas at pressure \(\mathrm{P}\) is (A) \(\mathrm{vP}\) (B) \(\mathrm{P}\) (C) \((\mathrm{p} / 2)\) (D) \((\mathrm{p} / \mathrm{v})\)

A uniform metal rod is used as a bas pendulum. If the room temperature rises by \(10^{\circ} \mathrm{C}\) and the efficient of line as expansion of the metal of the rod is, \(2 \times 10^{-6} 0_{\mathrm{c}}^{-1}\) what will have percentage increase in the period of the pendulum? (A) \(-2 \times 10^{-3}\) (B) \(1 \times 10^{-3}\) (C) \(-1 \times 10^{-3}\) (D) \(2 \times 10^{-3}\)

The work of \(62.25 \mathrm{KJ}\) is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by \(5^{\circ} \mathrm{C}\) The gas is $\mathrm{R}=8.3\\{\mathrm{~J} /(\mathrm{mole})\\}$ (A) triatomic (B) diatomic (C) monoatomic (D) a mixture of monoatomic and diatomic
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