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Chapter 13: Problem 48

A constant volume gas thermometer containing helium is immersed in boiling ammonia \(\left(-33^{\circ} \mathrm{C}\right)\) and the pressure is read once equilibrium is reached. The thermometer is then moved to a bath of boiling water \(\left(100.0^{\circ} \mathrm{C}\right) .\) After the manometer was adjusted to keep the volume of helium constant, by what factor was the pressure multiplied?

Short Answer

Expert verified
Answer: The pressure was multiplied by a factor of approximately 1.55.

Step by step solution

01

Understand the given information

We are given: - Temperature of boiling ammonia = \(-33^{\circ} \mathrm{C}\) - Temperature of boiling water = \(100.0^{\circ} \mathrm{C}\) - The volume of the gas remains constant.
02

Convert the temperatures to Kelvin scale

The temperatures given are in Celsius. We need to convert them to Kelvin to work with the ideal gas law. - Temperature of boiling ammonia in Kelvin = \((-33+273) K = 240 K\) - Temperature of boiling water in Kelvin = \((100+273) K = 373 K\)
03

Use the ideal gas law

The ideal gas law states: \(PV = nRT\) Since the volume and the amount of gas are constant, we can compare the pressures and temperatures as follows: - \(P_1/T_1 = P_2/T_2\) We need to find the factor by which the pressure was multiplied, which is \(P_2/P_1\)
04

Solve for the pressure factor

Now we can solve for the pressure factor (P2/P1) as follows: $$ \frac{P_2}{P_1}=\frac{T_2}{T_1}=\frac{373 \mathrm{K}}{240 \mathrm{K}} $$
05

Calculate the pressure factor

To find the factor by which the pressure was multiplied, divide the temperatures in Kelvin: $$ \frac{P_2}{P_1}=\frac{373}{240} \approx 1.55 $$ The pressure was multiplied by a factor of approximately 1.55.

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Most popular questions from this chapter

The mass of \(1 \mathrm{mol}\) of \(^{13} \mathrm{C}\) (carbon- 13 ) is $13.003 \mathrm{g}$. (a) What is the mass in \(u\) of one \(^{13} \mathrm{C}\) atom? (b) What is the mass in kilograms of one \(^{13} \mathrm{C}\) atom?
The coefficient of linear expansion of brass is \(1.9 \times 10^{-5}\) $^{\circ} \mathrm{C}^{-1} .\( At \)20.0^{\circ} \mathrm{C},$ a hole in a sheet of brass has an area of \(1.00 \mathrm{mm}^{2} .\) How much larger is the area of the hole at $30.0^{\circ} \mathrm{C} ?(\text { Wile tutorial: loop around the equator })$
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