Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem Statement:**
Calculate the density of N₂O at 95°C and 794 torr.
---
**Explanation:**
To solve this problem, we will use the ideal gas law and the molecular weight of nitrous oxide (N₂O). The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume of the gas,
- \( n \) is the number of moles of the gas,
- \( R \) is the ideal gas constant (\( 0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1} \) or other appropriate unit based on pressure and volume),
- \( T \) is the temperature in Kelvin.
Next, convert the temperature from Celsius to Kelvin and the pressure from torr to atm to use the ideal gas constant \( R \):
\[ T (K) = 95 + 273.15 = 368.15 \, K \]
\[ P (atm) = \frac{794 \, \text{torr}}{760 \, \text{torr/atm}} \approx 1.045 \, atm \]
To find the density, we also need the molar mass (M) of N₂O:
\[ \text{Molar Mass of } N_2O = (2 \times 14.01) + 16.00 = 44.02 \, \text{g/mol} \]
Using the ideal gas law and the relationship between density (ρ), molar mass (M), pressure (P), ideal gas constant (R), and temperature (T), we get the formula for density:
\[ \rho = \frac{MP}{RT} \]
Substitute the values into the equation:
\[ \rho = \frac{(44.02 \, \text{g/mol})(1.045 \, \text{atm})}{(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1}) (368.15 \, \text{K})} \]
Calculate the density:
\[ \rho \approx \frac{45.989}{30.225} \, \text{g/L} \approx 1.52 \, \text{g/L} \]
Therefore](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4963a1b7-231e-4415-aa15-591e301cddd8%2F83ac1aa0-636f-4c26-8fdd-5706142c8388%2Ft5jbof.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the density of N₂O at 95°C and 794 torr.
---
**Explanation:**
To solve this problem, we will use the ideal gas law and the molecular weight of nitrous oxide (N₂O). The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume of the gas,
- \( n \) is the number of moles of the gas,
- \( R \) is the ideal gas constant (\( 0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1} \) or other appropriate unit based on pressure and volume),
- \( T \) is the temperature in Kelvin.
Next, convert the temperature from Celsius to Kelvin and the pressure from torr to atm to use the ideal gas constant \( R \):
\[ T (K) = 95 + 273.15 = 368.15 \, K \]
\[ P (atm) = \frac{794 \, \text{torr}}{760 \, \text{torr/atm}} \approx 1.045 \, atm \]
To find the density, we also need the molar mass (M) of N₂O:
\[ \text{Molar Mass of } N_2O = (2 \times 14.01) + 16.00 = 44.02 \, \text{g/mol} \]
Using the ideal gas law and the relationship between density (ρ), molar mass (M), pressure (P), ideal gas constant (R), and temperature (T), we get the formula for density:
\[ \rho = \frac{MP}{RT} \]
Substitute the values into the equation:
\[ \rho = \frac{(44.02 \, \text{g/mol})(1.045 \, \text{atm})}{(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1}) (368.15 \, \text{K})} \]
Calculate the density:
\[ \rho \approx \frac{45.989}{30.225} \, \text{g/L} \approx 1.52 \, \text{g/L} \]
Therefore
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