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Chapter 17: Problem 66

Show that the coefficient of volume expansion of an ideal gas at constant pressure is the reciprocal of its kelvin temperature.

Short Answer

Expert verified
The coefficient of volume expansion (尾) at constant pressure for an ideal gas is equal to the reciprocal of the gas's absolute temperature. Hence \( 尾 = \frac{1}{T} \), with \( T \) as the temperature in Kelvins.

Step by step solution

01

Understanding Basic Principles

The first step is understanding the Ideal Gas Law, represented as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the universal gas constant, and \( T \) is the temperature in kelvin. Considering the gas at constant pressure means that \( P \) is a constant. The relationship to be shown involves the the coefficient of volume expansion (尾), which by definition is \( 尾 = \frac{1}{V}\frac{dV}{dT} \). This defines how the volume of a substance changes with temperature, with pressure held constant.
02

Derivation of the Relationship

Next, substitute \( V = \frac{nRT}{P} \) from the Ideal Gas Law into the equation for 尾. This gives \( 尾 = \frac{1}{\frac{nRT}{P}} \frac{d(\frac{nRT}{P})}{dT} \). Then find the derivative with respect to \( T \) to get \( 尾 = \frac{P}{nRT} \times nR.P^{-1} \). Simplifying this, cancelling out \( P \) and \( nR \), yields \( 尾 = \frac{1}{T} \). This shows that the volume expansion coefficient (尾) of an ideal gas at constant pressure is indeed the reciprocal of its absolute temperature \( T \) in kelvin.

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Most popular questions from this chapter

Your professor asks you to order a tank of argon gas for a lab experiment. You obtain a "type \(\mathrm{C}^{\prime \prime}\) gas cylinder with interior volume 6.88 L. The supplier claims it contains 45 mol of argon. You measure its pressure to be 14 MPa at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) Did you get what you paid for?

At high gas densities, the van der Waals equation modifies the ideal-gas law to account for nonzero molecular volume and for the van der Waals force that we discussed in Section \(17.1 .\) The van der Waals equation is $$\left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T$$ where \(a\) and \(b\) are constants that depend on the particular gas. For nitrogen \(\left(\mathrm{N}_{2}\right), a=0.14 \mathrm{Pa} \cdot \mathrm{m}^{6} / \mathrm{mol}^{2}\) and \(b=3.91 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol}\) For 1.000 mol of \(\mathrm{N}_{2}\) at 10.00 atm pressure, confined to a volume of \(2.000 \mathrm{L},\) find the temperatures predicted (a) by the ideal-gas law and (b) by the van der Waals equation.

Your roommate claims that ice and snow must be at \(0^{\circ} \mathrm{C}\). Is that true?

The atmospheres of relatively low-mass planets like Earth don't contain much hydrogen (H \(_{2}\) ), while more massive planets like Jupiter have considerable atmospheric hydrogen. What factors might account for the difference?

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