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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

In a container of negligible heat capacity, \(200 \mathrm{~g}\) 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 \(1 \mathrm{~atm} .\) Amount of the Sm left in the system, is equal to (A) \(16.7 \mathrm{~g}\) (B) \(8.4 \mathrm{~g}\) (C) \(12 \mathrm{~g}\) (D) \(0 \mathrm{~g}\) Copyright () StemEZ.com. All rights reserved.

Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

A monoatomic ideal gas, intially at temperature \(1_{1}\) is enclosed in a cylinders fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature \(\mathrm{T}_{2}\) by releasing the piston suddenly If \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\) the lengths of the gas column before and after expansion respectively, then then \(\left(\mathrm{T}_{1} / \mathrm{T}_{2}\right)\) is given by (A) \(\left\\{\mathrm{L}_{1} / \mathrm{L}_{2}\right\\}^{(2 / 3)}\) (B) \(\left\\{\mathrm{L}_{2} / \mathrm{L}_{1}\right\\}^{(2 / 3)}\) (C) \(\left\\{\mathrm{L}_{1} / \mathrm{L}_{2}\right\\}\) (D) \(\left\\{\mathrm{L}_{2} / \mathrm{L}_{1}\right\\}\)

If a heat engine absorbs \(2 \mathrm{KJ}\) heat from a heat source and release \(1.5 \mathrm{KJ}\) heat into cold reservoir, then its efficiency is (A) \(0.5 \%\) (B) \(75 \%\) (C) \(25 \%\) (D) \(50 \%\)

If for a gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=1.67$, this gas is made up to molecules which are (A) diatomic (B) Polytomic (C) monoatomic (D) mixnese of diatomic and polytomic molecules
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