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Chapter 8: Problem 1178

What is the relationship Pressure and temperature for an ideal gas undergoing adiabatic Change. (A) \(\mathrm{PT}^{\gamma}=\) Const (B) \(\mathrm{PT}^{-1+\gamma}=\) Const (C) \(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\) Const (D) \(\mathrm{P}^{\gamma-1} \mathrm{~T}^{\gamma}=\) Const

Short Answer

Expert verified
The relationship between Pressure and temperature for an ideal gas undergoing adiabatic change is given by: \(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma} = \text{Const}\) (Option C).

Step by step solution

01

Remember the adiabatic equation for an ideal gas

The adiabatic equation for an ideal gas is \(PV^{\gamma} = \text{constant}\), where P is pressure, V is volume, \(\gamma\) (gamma) is the specific heat ratio or adiabatic index (\(\frac{C_p}{C_v}\)), and \(PV^{\gamma}\) is constant during the adiabatic process.
02

Eliminate volume (V) term from adiabatic equation

To eliminate the volume (V) term, we can use the ideal gas equation, \(PV=nRT\), where n is the number of moles of the gas and R is the universal gas constant. Divide both sides by P: \(V = \frac{nRT}{P}\) Now substitute this expression for V in the adiabatic equation: \(P\left(\frac{nRT}{P}\right)^{\gamma} = \text{constant}\)
03

Simplify the equation

Distribute the \(\gamma\) exponent: \(P\left(\frac{n^{\gamma}R^{\gamma}T^{\gamma}}{P^{\gamma}}\right) = \text{constant}\) Simplify the equation by multiplying P: \(P^{1-\gamma}T^{\gamma} = \frac{n^{\gamma}R^{\gamma}}{\text{constant}}\) Since we need to find the relationship between P and T, both sides of the equation are constants, so it can be expressed as: \(P^{1-\gamma}T^{\gamma} = \text{constant}\)
04

Compare with given options

Based on our derived relationship between pressure and temperature for an ideal gas undergoing adiabatic change, the correct option is: (C) \(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma} = \text{Const}\)

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Most popular questions from this chapter

Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(B\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

In an isothermal reversible expansion, if the volume of \(96 \mathrm{~J}\) of oxygen at \(27^{\circ} \mathrm{C}\) is increased from 70 liter to 140 liter, then the work done by the gas will be (A) \(300 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (C) \(2.3 \times 900 \mathrm{R} \log _{10} 2\) (D) \(100 \mathrm{R} \log _{10}^{(2)}\)

If du represents the increase in internal energy of a thermodynamic system and dw the work done by the system, which of the following statement is true? (A) \(\mathrm{du}=\mathrm{dw}\) in isothermal process (C) \(\mathrm{du}=-\mathrm{dw}\) in an aidabatic process (B) \(\mathrm{du}=\mathrm{dw}\) in aidabatic process (D) \(\mathrm{du}=-\mathrm{dw}\) in an isothermal process

For free expansion of the gas which of the following is true ? (A) \(\mathrm{Q}=0, \mathrm{~W}>0\) and \(\mu \operatorname{Eint}=-\mathrm{W}\) (B) \(\mathrm{W}=0, \mathrm{Q}>0\) and \(\Delta\) Eint \(=\mathrm{Q}\) (C) \(\mathrm{W}>0, \mathrm{Q}<0\) and \(\Delta\) Eint \(=0\) (D) \(\mathrm{Q}=\mathrm{W}=0\) and \(\Delta \operatorname{Eint}=0\)

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) \(20 \mathrm{KJ}\)
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