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At \(\mathrm{O}^{\circ} \mathrm{C}\) the density of a fixed mass of a gas divided by pressure is \(\mathrm{x} .\) At \(100^{\circ} \mathrm{C}\), the ratio will be (A) \(\mathrm{x}\) (B) \([(273) /(373)] \mathrm{x}\) (C) \([(373) /(273)] \mathrm{x}\) (D) \([(100) /(273)] \mathrm{x}\)

Short Answer

Expert verified
The short answer is: \( \frac{蟻_2}{P_2} = \frac{373}{273}x \) So, the correct answer is: (C) \([(373) /(273)] \mathrm{x}\)

Step by step solution

01

Recall the ideal gas law

The ideal gas law can be written as: \( PV = nRT \) Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature (in Kelvin)
02

Express density in terms of the ideal gas law

Density (蟻) can be defined as mass (m) divided by volume (V). For a fixed mass of gas, we can express the number of moles (n) as mass (m) divided by the molar mass (M). Thus, n = m/M. Now we can rewrite the ideal gas law in terms of mass (m), molar mass (M), and density (蟻): \( P(V) = \frac{m}{M} RT \) Now, rearrange for the density: \( 蟻 = \frac{m}{V} = \frac{P M}{RT} \)
03

Divide both sides of the equation by pressure (P)

We want to get the density divided by pressure (蟻/P). Therefore, divide the equation from step 2 by P. \( \frac{蟻}{P} = \frac{M}{RT} \)
04

Plug in the values for 0掳C and 100掳C

Now, let's plug in the values for T = 0掳C and T = 100掳C and find the ratios. T1 = 0掳C = 273 K T2 = 100掳C = 373 K At 0掳C: \( \frac{蟻_1}{P_1} = \frac{M}{R 脳 273} \) , where 蟻1/P1 corresponds to x. At 100掳C: \( \frac{蟻_2}{P_2} = \frac{M}{R 脳 373} \) Now, let's find the ratio of these expressions. Divide the expression at 100掳C by the expression at 0掳C: \( \frac{\frac{蟻_2}{P_2}}{\frac{蟻_1}{P_1}} = \frac{\frac{M}{R 脳 373}}{\frac{M}{R 脳 273}} \)
05

Simplify the expression

Cancel out M and R in the expression above: \( \frac{\frac{蟻_2}{P_2}}{x} = \frac{373}{273} \) Now, isolate the term at 100掳C: \( \frac{蟻_2}{P_2} = \frac{373}{273}x \) So, the correct answer is: (C) \([(373) /(273)] \mathrm{x}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density and Pressure Relationship
The relationship between density and pressure in gases is closely linked to the ideal gas law. To understand this, let's start with the basics. Density \((\rho)\) is defined as the mass \((m)\) of an object divided by its volume \((V)\). For gases, this means that:
\[ \rho = \frac{m}{V} \]
Now, using the ideal gas law:
\[ PV = nRT \]
where \(P\) stands for pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. The number of moles \(n\) can also be expressed as the mass \(m\) divided by the molar mass \(M\). Therefore, \(n = \frac{m}{M}\). Substituting this into the ideal gas law formula gives us:
\[ PV = \frac{m}{M}RT \]
Rearranging the equation to solve for density and dividing by pressure \(P\), we develop the equation:
\[ \frac{\rho}{P} = \frac{M}{RT} \]
This equation demonstrates that density divided by pressure is inversely proportional to temperature, if the molar mass is held constant. As a result, as the temperature increases, \(\rho/P\) decreases, assuming pressure remains constant. This derivation is critical for predicting how gases behave under various conditions.
Temperature Conversion to Kelvin
Temperature is crucial in gas behavior analysis, often requiring conversion to the Kelvin scale, as this relates directly to the absolute energy of molecules. The Kelvin scale starts at absolute zero, where theoretically molecular motion stops. To convert Celsius to Kelvin, you simply add 273.15 (often approximated as 273 for simplicity in problem solving).
For example, to convert \(0^{\circ}C\) to Kelvin:
\[ T(\text{K}) = 0 + 273 = 273 \text{ K} \]
And to convert \(100^{\circ}C\):
\[ T(\text{K}) = 100 + 273 = 373 \text{ K} \]
These conversions are essential because gas laws, like the ideal gas law, require absolute temperatures measured in Kelvin to accurately predict how gases behave under different conditions.
Gas Laws in Thermodynamics
The gas laws serve as fundamental principles in thermodynamics, crucial for understanding the behavior of gases. The most prominent among them is the Ideal Gas Law represented as:
\[ PV = nRT \]
This law is a unifying equation, encompassing the behavior described by more specific gas laws such as Boyle鈥檚 Law, Charles鈥檚 Law, and Avogadro鈥檚 Law.
  • **Boyle's Law:** At constant temperature, the pressure and volume of a gas are inversely proportional. \(P_1V_1 = P_2V_2\)
  • **Charles's Law:** At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
  • **Avogadro's Law:** For a given temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. \(V = kn\)

When combined, these laws help predict changes in state or conditions for a gas. By understanding these principles, we gain valuable insights into thermal systems, addressing real-world problems ranging from weather forecasting to engine designs.

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Most popular questions from this chapter

The rms. speed of the molecules of a gas in a vessel is \(400 \mathrm{~ms}^{-1}\). If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be (A) \(800 \mathrm{~ms}^{-1}\) (B) \(200 \mathrm{~ms}^{-1}\) (C) \(400 \sqrt{2} \mathrm{~ms}^{-1}\) (D) \(400 \mathrm{~ms}^{-1}\)

To what temperature should the hydrogen at room temperature \(\left(27^{\circ} \mathrm{C}\right)\) be heated at constant pressure so that the rms velocity of its molecule becomes double of its previous value (A) \(927^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(108^{\circ} \mathrm{C}\) (D) \(1200^{\circ} \mathrm{C}\)

The average translational energy and \(\mathrm{rms}\) speed of molecules in sample of oxygen gas at \(300 \mathrm{~K}\) are \(6.21 \times 10^{-21} \mathrm{~J}\) and \(484 \mathrm{~m} / \mathrm{s}\) respectively. The corresponding values at \(600 \mathrm{~K}\) are nearly (assuming ideal gas behavior) (A) \(6.21 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (B) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\) (C) \(12.42 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (D) \(8.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\)

Under constant temperature, graph between \(\mathrm{p}\) and \((1 / \mathrm{V})\) is (A) Hyperbola (B) Circle (C) Parabola (D) Straight line

Calculate the temperature at which \(\mathrm{rms}\) velocity of \(\mathrm{SO}_{2}\) molecules is the same as that of \(\mathrm{O}_{2}\) molecules at \(27^{\circ} \mathrm{C}\). Molecular weights of Oxygen and \(\mathrm{SO}_{2}\) are \(32 \mathrm{~g}\) and \(64 \mathrm{~g}\) respectively (A) \(327^{\circ} \mathrm{C}\) (B) \(327 \mathrm{~K}\) (C) \(127^{\circ} \mathrm{C}\) (D) \(227^{\circ} \mathrm{C}\)

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