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Under the identical conditions of temper...

Under the identical conditions of temperature, the density of a gas X is two times to that of gas Y while molecular mass of gas Y is three times that of X. Calculate the ratio of pressure of X and Y.

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To solve the problem, we will use the Ideal Gas Law and the relationships between density, molecular mass, and pressure. Let's break it down step by step. ### Step 1: Understand the Given Information - The density of gas X (denoted as \( \rho_X \)) is twice that of gas Y (denoted as \( \rho_Y \)): \[ \rho_X = 2 \rho_Y \] - The molecular mass of gas Y (denoted as \( M_Y \)) is three times that of gas X (denoted as \( M_X \)): \[ M_Y = 3 M_X \] ### Step 2: Apply the Ideal Gas Law The Ideal Gas Law is given by: \[ PV = nRT \] Where: - \( P \) = Pressure - \( V \) = Volume - \( n \) = Number of moles - \( R \) = Universal gas constant - \( T \) = Temperature Since the temperature and the universal gas constant are constant for both gases, we can express the pressure in terms of density and molecular mass. ### Step 3: Relate Pressure, Density, and Molecular Mass From the Ideal Gas Law, we can express the number of moles \( n \) in terms of density and molecular mass: \[ n = \frac{m}{M} \] Where \( m \) is the mass of the gas. The volume \( V \) can also be expressed as: \[ V = \frac{m}{\rho} \] Substituting \( n \) and \( V \) into the Ideal Gas Law gives: \[ P \left(\frac{m}{\rho}\right) = \frac{m}{M} RT \] Rearranging gives: \[ P = \frac{\rho RT}{M} \] ### Step 4: Calculate the Ratio of Pressures Now, we can find the ratio of pressures for gases X and Y: \[ \frac{P_X}{P_Y} = \frac{\rho_X \cdot R \cdot T / M_X}{\rho_Y \cdot R \cdot T / M_Y} = \frac{\rho_X}{\rho_Y} \cdot \frac{M_Y}{M_X} \] ### Step 5: Substitute the Known Values From the information given: - \( \frac{\rho_X}{\rho_Y} = 2 \) - \( \frac{M_Y}{M_X} = 3 \) Substituting these values into the ratio of pressures: \[ \frac{P_X}{P_Y} = 2 \cdot 3 = 6 \] ### Final Answer Thus, the ratio of the pressures of gas X to gas Y is: \[ \frac{P_X}{P_Y} = 6 \] ---

To solve the problem, we will use the Ideal Gas Law and the relationships between density, molecular mass, and pressure. Let's break it down step by step. ### Step 1: Understand the Given Information - The density of gas X (denoted as \( \rho_X \)) is twice that of gas Y (denoted as \( \rho_Y \)): \[ \rho_X = 2 \rho_Y \] - The molecular mass of gas Y (denoted as \( M_Y \)) is three times that of gas X (denoted as \( M_X \)): ...
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