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Chapter 6: Problem 136

Which of the following choices represents the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1.5 atm? (a) \((298 \times 1.5 / 273) \times 22.4 \mathrm{L} ;\) (b) \(22.4 \mathrm{L}\) (c) \((273 \times 1.5 / 298) \times 22.4 \mathrm{L}\) (d) \([298 /(273 \times 1.5)] \times 22.4 \mathrm{L}\) (e) \([273 /(298 \times 1.5)] \times 22.4 \mathrm{L}\)

Short Answer

Expert verified
The molar volume of an ideal gas at 25°C and 1.5 atm is approximately 20.7 L or, therefore, the closest choice is (c) \((273*1.5/298) \times 22.4\) L.

Step by step solution

01

Convert Temperature to Kelvin

First, we'll need to convert the given temperature from Celsius to Kelvin. Add 273.15 to the Celsius temperature to do this: \(25^{\circ}C + 273.15 = 298.15 K\). For simplicity, we will round this to 298 K.
02

Adjust Molar Volume

We know that at 1 atm and 273 K, the molar volume of an ideal gas is approximately 22.4 L. But here the temperature is 298 K and the pressure is 1.5 atm. This will affect the molar volume proportionally. It will increase with the temperature and decrease with the pressure. Therefore, we need to adjust the 22.4 L value to reflect the change in temperature and pressure. The correct formula for adjustment should be \((273*1.5/298) \times 22.4\) L.
03

Calculate the New Molar Volume

Now plug in the known values to calculate the new molar volume under these conditions: \((273*1.5/298) \times 22.4\) L = \(20.7\) L. So, the molar volume of an ideal gas at 25°C and 1.5 atm is approximately \(20.7\) L.

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Most popular questions from this chapter

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).

If 3.0 L of oxygen gas at \(177^{\circ} \mathrm{C}\) is cooled at constant pressure until the volume becomes \(1.50 \mathrm{L}\), then what is the final temperature?

How many liters of \(\mathrm{H}_{2}(\mathrm{g})\) at STP are produced per gram of \(\mathrm{Al}(\mathrm{s})\) consumed in the following reaction? \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

The equation \(d / P=M / R T,\) which can be derived from equation \((6.14),\) suggests that the ratio of the density \((d)\) to pressure (P) of a gas at constant temperature should be a constant. The gas density data at the end of this question were obtained for \(\mathrm{O}_{2}(\mathrm{g})\) at various pressures at \(273.15 \mathrm{K}\) (a) Calculate values of \(d / P,\) and with a graph or by other means determine the ideal value of the term \(d / P\) for \(\mathrm{O}_{2}(\mathrm{g})\) at \(273.15 \mathrm{K}\) [Hint: The ideal value is that associated with a perfect (ideal) gas.] (b) Use the value of \(d / P\) from part (a) to calculate a precise value for the atomic mass of oxygen, and compare this value with that listed on the inside front cover. $$\begin{array}{lllll} P, \mathrm{mmHg}: & 760.00 & 570.00 & 380.00 & 190.00 \\ d, \mathrm{g} / \mathrm{L}: & 1.428962 & 1.071485 & 0.714154 & 0.356985 \end{array}$$

A 2.650 g sample of a gaseous compound occupies \(428 \mathrm{mL}\) at \(24.3^{\circ} \mathrm{C}\) and \(742 \mathrm{mmHg} .\) The compound consists of \(15.5 \%\) C \(, 23.0 \%\) Cl, and \(61.5 \%\) F, by mass. What is its molecular formula?
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