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Chapter 6: Problem 116

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l} 313 \text { g He } \\ 1.00 \text { atm } \\ \text { 1910. L } \\ \text { Molar Heat Capacity }=20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol} \end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{~L}\), with the pressure remaining constant. Determine \(q, w\), and \(\Delta E\) (in \(\mathrm{kJ}\) ) for the compression of the balloon.

Short Answer

Expert verified
For the compression of the helium-filled balloon, the calculated values are: heat, \(q = -68.1 \: kJ\); work, \(w = 267 \: kJ\); and change in internal energy, \(\Delta E = 198.9 \: kJ\).

Step by step solution

01

Calculate the moles of helium

Given the mass of helium (313g) and the molar mass of helium (4.00 g/mol), we can calculate the moles (n) of helium in the balloon: \[ n = \frac{mass_{He}}{MolarMass_{He}} \] \[ n = \frac{313}{4.00} = 78.25 mol \]
02

Find the initial and final temperature

We will use the ideal gas law for both initial and final states to find the initial and final temperatures: \( P_1V_1 = nRT_1 \) and \(P_2V_2 = nRT_2 \) Given: \( P_1 = P_2; V_1 = 1910 L; V_2 = 1643 L; n = 78.25 mol; R = 0.0821 \frac{L.atm}{K.mol} \) \[ T_1 = \frac{P_1V_1}{nR} \] \[ T_1 = \frac{(1.00)(1910)}{(78.25)(0.0821)} = 298.6 K \] The temperature is decreased by \(41.6^{\circ}C\), so: \[ \Delta T = -41.6 K \] \[ T_2 = T_1 + \Delta T = 298.6 K - 41.6 K = 257.0 K \]
03

Calculate heat (q)

We will use the formula: \[ q = n\times C_{p} \times \Delta T \] Where, \( q \) is the heat, \( n \) is the number of moles, \( C_{p} \) is the molar heat capacity, and \( \Delta T \) is the change in temperature. \[ q = (78.25) \times (20.8) \times (-41.6) = -68,130 J \] Convert this to kJ: \[ q = -68.1 kJ \]
04

Calculate work (w)

The formula for work (w) is given by: \[ w = -P \times \Delta V \] Where, \(P\) is the pressure, and \(\Delta V\) is the change in volume. \[ \Delta V = V_2 - V_1 = 1643 - 1910 = -267 L \] \[ w = -(1.00) \times (-267) = 267 kJ \]
05

Calculate change in internal energy (ΔE)

We can calculate the change in internal energy (ΔE) using the formula: \[ \Delta E = q + w \] \[ \Delta E = (-68.1) + (267) = 198.9 kJ \] The values of heat (q), work (w), and change in internal energy (ΔE) for the compression of the balloon are: \[ q = -68.1 \: kJ \] \[ w = 267 \: kJ \] \[ \Delta E = 198.9 \: kJ \]

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Most popular questions from this chapter

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