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Chapter 10: Problem 118

Consider the data for substance \(X\) given in Exercise 117 . When the temperature of 1.000 mole of \(X(g)\) is lowered from \(100.0^{\circ} \mathrm{C}\) to form \(\mathrm{X}(l)\) at \(50.0^{\circ} \mathrm{C}, 28.75 \mathrm{kJ}\) of heat is released. Calculate the specific heat capacity of \(\mathrm{X}(g)\)

Short Answer

Expert verified
The specific heat capacity of substance X in the gaseous state is \(575 J/(mol * K)\).

Step by step solution

01

Convert temperature to Kelvin

Since specific heat capacity depends on temperature, we need to make sure we work in Kelvin. To convert temperature from Celsius to Kelvin, we add 273.15. Thus, the initial and final temperatures in Kelvin are: Initial temperature (T1) = 100.0°C + 273.15 = \(373.15 K\) Final temperature (T2) = 50.0°C + 273.15 = \(323.15 K\)
02

Calculate the change in temperature

We'll now find the change in temperature (ΔT). To do this, subtract the final temperature from the initial temperature: ΔT = T2 - T1 = \(323.15 K - 373.15 K = -50 K\)
03

Solve for the specific heat capacity

We can now use the formula q = n * C * ΔT to solve for the specific heat capacity (C) of X in the gaseous state. We are given the values for q, n, and ΔT: q = -28.75 kJ (heat released, so negative) n = 1 mole ΔT = -50 K Remember to convert q to J by multiplying by 1000: q = -28750 J Now, plug in the provided values and solve for C: \(-28750 J = 1 * C * -50 K\) First, cancel the negative signs: \(28750 J = 1 * C * 50 K\) Now, divide by 50 and 1 to get the specific heat capacity: \(C = \dfrac{28750 J}{1 * 50 K} = 575 J/(mol * K)\)
04

Present the answer

The specific heat capacity of substance X in the gaseous state is \(575 J/(mol * K)\).

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Most popular questions from this chapter

Consider the following formulas for n-pentane and neopentane: Both compounds have the same overall formula $\left(\mathrm{C}_{5} \mathrm{H}_{12}, \text { molar }\right.\( mass \)=72.15 \mathrm{g} / \mathrm{mol} ),\( yet \)n\( -pentane boils at \)36.2^{\circ} \mathrm{C}$ whereas neopentane boils at \(9.5^{\circ} \mathrm{C} .\) Rationalize the differences in the boiling points between these two nonpolar compounds.

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