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Chapter 10: Problem 33

Complete the following table for an ideal gas: $$\begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & ? \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & ? \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & ? \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ ? \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array}$$

Short Answer

Expert verified
The completed table is: $$ \begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & 728.24 \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & 0.2536 \mathrm{~mol} & 300 \mathrm{~K} \\ 101.33 \mathrm{kPa} & 8.21 \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\ 24.7 \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array} $$

Step by step solution

01

Row 1: Find T

Given values: P = 303.98 kPa, V = 3.00 L, n = 1.500 mol, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for T. Using ideal gas law equation: 303.98 kPa × 3.00 L = 1.500 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × T Solve for T: T = \(\frac{303.98 \times 3.00}{1.500 \times 0.0831}\) = 728.24 K
02

Row 2: Find n

Given values: P = 50.663 kPa, V = 0.750 L, T = 300 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for n. Using ideal gas law equation: 50.663 kPa × 0.750 L = n × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 300 K Solve for n: n = \(\frac{50.663 \times 0.750}{0.0831 \times 300}\) = 0.2536 mol
03

Row 3: Find V

Given values: P = 101.33 kPa, n = 3.333 mol, T = 300 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for V. Using the ideal gas law equation: 101.33 kPa × V = 3.333 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 300 K Solve for V: V = \(\frac{3.333 \times 0.0831 \times 300}{101.33}\) = 8.21 L
04

Row 4: Find P

Given values: V = 0.750 L, n = 0.750 mol, T = 298 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for P. Using ideal gas law equation: P × 0.750 L = 0.750 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 298 K Solve for P: P = \(\frac{0.750 \times 0.0831 \times 298}{0.750}\) = 24.7 kPa The completed table is: $$\begin{array}{llll} P & V & n & T \\\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & 728.24 \mathrm{~K} \\\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & 0.2536 \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & 8.21 \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ 24.7 \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\\ \hline \end{array}$$

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Most popular questions from this chapter

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4},\) is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is \(1 \mathrm{ppb}\) (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and 101.3 kPa pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m} ?\)

Consider the following gases, all at STP: $\mathrm{Ne}, \mathrm{SF}_{6}, \mathrm{~N}_{2}, \mathrm{CH}_{4}$. (a) Which gas is most likely to depart from the assumption of the kinetic- molecular theory that says there are no attractive or repulsive forces between molecules? (b) Which one is closest to an ideal gas in its behavior? (c) Which one has the highest root-mean-square molecular speed at a given temperature? (d) Which one has the highest total molecular volume relative to the space occupied by the gas? (e) Which has the highest average kinetic- molecular energy? (f) Which one would effuse more rapidly than $\mathrm{N}_{2} ?\( (g) Which one would have the largest van der Waals \)b$ parameter?

Consider the combustion reaction between \(1.00 \mathrm{~L}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(500 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the rootmean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; \(\mathbf{d}\) ) the total number of collisions of molecules with walls per second.

As discussed in the "Chemistry Put to Work" box in Section 10.8 , enriched uranium can be produced by effusion of gaseous \(\mathrm{UF}_{6}\) across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, \(\mathrm{U}(\mathrm{g})\). Calculate the ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\), and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.
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