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Chapter 5: Problem 61

A gas sample containing 1.50 moles at \(25^{\circ} \mathrm{C}\) exerts a pressure of 400. torr. Some gas is added to the same container and the temperature is increased to 50\(\cdot^{\circ} \mathrm{C}\). If the pressure increases to 800. torr, how many moles of gas were added to the container? Assume a constant-volume container.

Short Answer

Expert verified
Approximately 0.7537 moles of gas were added to the container.

Step by step solution

01

Convert temperatures to Kelvin

Since the given temperatures are in Celsius, we first need to convert them to Kelvin. So, add 273.15 to each given temperature: T1 = 25 + 273.15 = 298.15 K T2 = 50 + 273.15 = 323.15 K
02

Convert pressure to atm

The given pressures are in torr unit, we need to convert them to atm unit, using the fact that 1 atm = 760 torr: P1 = 400 torr * (1 atm / 760 torr) = 0.5263 atm P2 = 800 torr * (1 atm / 760 torr) = 1.0526 atm
03

Use the Ideal Gas Law

Since the volume and gas constant R are constant, we can work with the ratios of pressure, moles, and temperature. \[\frac{P_1V}{n_1RT_1}=\frac{P_2V}{n_2RT_2}\] \[\frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}\]
04

Plug in the known values

We can now plug in the known values for P1, n1, T1, P2, and T2, and solve for n2 (the moles after adding gas). \[\frac{0.5263}{1.50 \times 298.15} = \frac{1.0526}{n_2 \times 323.15}\]
05

Solve for n2

Solving for n2, we get: \[n_2 = \frac{1.0526 \times 1.50 \times 298.15 \times 323.15}{0.5263 \times 298.15}\] \[n_2 = 2.2537\]
06

Find the difference in moles

Now that we have n2 (the number of moles after adding gas), we can find the difference in the moles, which represents the moles of gas added to the container. moles_added = n2 - n1 moles_added = 2.2537 - 1.50 moles_added = 0.7537 moles So, approximately 0.7537 moles of gas were added to the container.

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