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Chapter 5: Problem 57

Consider two separate gas containers at the following conditions: $$ \begin{array}{|ll|} \text { Container A } & \text { Container B } \\ \text { Contents: } \mathrm{SO}_{2}(g) & \text { Contents: unknown gas } \\ \text { Pressure }=P_{\mathrm{A}} & \text { Pressure }=P_{\mathrm{B}} \\ \text { Moles of gas }=1.0 \mathrm{~mol} & \text { Moles of gas }=2.0 \mathrm{~mol} \\ \text { Volume }=1.0 \mathrm{~L} & \text { Volume }=2.0 \mathrm{~L} \\ \text { Temperature }=7^{\circ} \mathrm{C} & \text { Temperature }=287^{\circ} \mathrm{C} \\ \hline \end{array} $$ How is the pressure in container \(\mathrm{B}\) related to the pressure in container \(\mathrm{A}\) ?

Short Answer

Expert verified
In conclusion, the pressure in Container B is twice the pressure in Container A, or \(P_B = 2 P_A\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. The conversion is done using the formula K = °C + 273.15. For Container A: Temperature (in Kelvin) = \(7 + 273.15\), \(T_A = 280.15 K\) For Container B: Temperature (in Kelvin) = \(287 + 273.15\), \(T_B = 560.15 K\)
02

Applying the Ideal Gas Law to Container A and Container B

Apply the Ideal Gas Law (\(PV=nRT\)) to both containers: For Container A: \(P_A \times 1.0 L = 1.0 mol \times R \times 280.15 K\) For Container B: \(P_B \times 2 L = 2.0 mol \times R \times 560.15 K\)
03

Solve for the pressure in Container B in terms of pressure in Container A

Our main goal in this step is to determine the relationship between the pressure in Container B and the pressure in Container A. Notice that R is a constant, so we can write the equations as: \(P_A = R \times 280.15 K\) \(2 P_B = 2 R \times 560.15 K\) Divide the second equation by 2: \(P_B = R \times 560.15 K\) Now, we need to relate the equations to each other. Divide the second equation by the first equation: \(\frac{P_B}{P_A} = \frac{R \times 560.15 K}{R \times 280.15 K}\) Since R is a constant, it cancels out: \(\frac{P_B}{P_A} = \frac{560.15}{280.15}\)
04

Calculate the relationship between the pressures

We can now determine the relationship between the pressures: \(\frac{P_B}{P_A} = \frac{560.15}{280.15} = 2.0\)
05

Conclusion

The pressure in Container B is twice the pressure in Container A. That is, \(P_B = 2 P_A\).

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Most popular questions from this chapter

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}\) ?

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