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Chapter 13: Problem 50

An ideal gas that occupies \(1.2 \mathrm{m}^{3}\) at a pressure of $1.0 \times 10^{5} \mathrm{Pa}\( and a temperature of \)27^{\circ} \mathrm{C}$ is compressed to a volume of \(0.60 \mathrm{m}^{3}\) and heated to a temperature of \(227^{\circ} \mathrm{C} .\) What is the new pressure?

Short Answer

Expert verified
Answer: The new pressure after the compression and heating process is approximately \(2.0\times10^{5}\,\mathrm{Pa}\).

Step by step solution

01

Write the Ideal Gas Law for the initial and final states

We are given information on the initial and final states of the ideal gas: initial volume (V1), initial pressure (P1), initial temperature (T1), final volume (V2), and final temperature (T2). We can write the Ideal Gas Law for these two states as: $$ P_{1}V_{1}=nRT_{1} \quad \text{and} \quad P_{2}V_{2}=nRT_{2} $$ Since the number of gas particles (n) and the gas constant (R) don't change, we can equate the two expressions.
02

Equate and simplify the Ideal Gas Law expressions for the initial and final states

We can now equate the expressions for the initial and final states and solve for the new pressure, P2: $$ \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} $$
03

Plug in the given values and convert temperatures to Kelvin

Before we plug in the given values, make sure to convert the temperatures from Celsius to Kelvin by adding 273.15: $$ T_{1}=27^{\circ}\mathrm{C}+273.15=300.15\,\mathrm{K} \\ T_{2}=227^{\circ}\mathrm{C}+273.15=500.15\,\mathrm{K} $$ Now insert the given values into the equation: $$ \frac{1.0\times10^{5}\mathrm{Pa}\cdot1.2\mathrm{m}^{3}}{300.15\,\mathrm{K}}=\frac{P_{2}(0.6\mathrm{m}^{3})}{500.15\,\mathrm{K}} $$
04

Solve for the new pressure, P2

To find the new pressure, P2, we can now solve the equation: $$ P_{2}=\frac{1.0\times10^{5}\mathrm{Pa}\cdot1.2\mathrm{m}^{3}\cdot500.15\,\mathrm{K}}{300.15\,\mathrm{K}\cdot0.6\mathrm{m}^{3}} $$ Calculating the result gives us: $$ P_{2}\approx2.0\times10^{5}\,\mathrm{Pa} $$ The new pressure after the compression and heating process is approximately \(2.0\times10^{5}\,\mathrm{Pa}\).

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Most popular questions from this chapter

The volume of a solid cube with side \(s_{0}\) at temperature \(T_{0}\) is \(V_{0}=s_{0}^{3} .\) Show that if \(\Delta s < s_{0},\) the change in volume \(\Delta V\) due to a change in temperature \(\Delta T\) is given by $$ \frac{\Delta V}{V_{0}}=3 \alpha \Delta T $$ and therefore that \(\beta=3 \alpha .\) (Although we derive this relation for a cube, it applies to a solid of any shape.)

A long, narrow steel rod of length \(2.5000 \mathrm{m}\) at \(25^{\circ} \mathrm{C}\) is oscillating as a pendulum about a horizontal axis through one end. If the temperature changes to \(0^{\circ} \mathrm{C}\) what will be the fractional change in its period?
A steel sphere with radius \(1.0010 \mathrm{cm}\) at \(22.0^{\circ} \mathrm{C}\) must slip through a brass ring that has an internal radius of $1.0000 \mathrm{cm}$ at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at \(22.0^{\circ} \mathrm{C},\) can just slip through?
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?
(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
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