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Chapter 19: Problem 27

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

Short Answer

Expert verified
The entropy change (螖S) for the isothermal process where the volume of helium gas is increased from 2.00 L to 5.00 L at 300 K, is approximately \(2.071~J/K\).

Step by step solution

01

Convert temperature to Kelvin

In thermodynamics, it's crucial to use the Kelvin temperature scale. To convert the given temperature (27掳C) to Kelvin, we need to add 273: T = 27掳C + 273 = \(300 K\) Now, we have the temperature in Kelvin and are ready to proceed.
02

Use the formula for entropy change

We will use the formula for entropy change in an isothermal process for an ideal gas: 螖S = nR * ln(V2/V1) We have: - n = 0.100 mol - R = 8.314 J/mol K (gas constant) - V1 = 2.00 L - V2 = 5.00 L
03

Calculate the entropy change

Plug the values into the 螖S formula and calculate the entropy change: 螖S = (0.100 mol) * (8.314 J/mol K) * ln(\(5.00 L / 2.00 L\)) 螖S 鈮� (0.100 mol) * (8.314 J/mol K) * ln(2.5) 螖S 鈮� \(2.071~J/K\) The entropy change for the process is approximately \(2.071~J/K\).

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Most popular questions from this chapter

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K\), at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

Thenormal boiling point of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(64.7^{\circ} \mathrm{C}\), and its molar enthalpy of vaporization is \(\Delta H_{\mathrm{vap}}=\) \(71.8 \mathrm{~kJ} / \mathrm{mol} .\) (a) When \(\mathrm{CH}_{3} \mathrm{OH}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00\) mol of \(\mathrm{CH}_{3} \mathrm{OH}(t)\) is vaporized at \(64.7^{\circ} \mathrm{C}\).

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \mathrm{n} K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?
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