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Google Chrome isn't your default browser Set as default Question 11 of 12 What is the pressure in atm of a 5.00 L tank with 7.95 moles of oxygen at 39.3 °C?

Google Chrome isn't your default browser Set as default Question 11 of 12 What is the pressure in atm of a 5.00 L tank with 7.95 moles of oxygen at 39.3 °C?

Chemistry
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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**Question 11 of 12** **Chemistry Problem:** *What is the pressure in atm of a 5.00 L tank with 7.95 moles of oxygen at 39.3 °C?* **Solution Interface:** - The diagram on the right shows a digital calculator interface. - The interface includes a numerical keypad with digits (1 to 9, 0), a decimal point (.), a button for exponentiation (x 10^), and operations for clearing (C) and deleting entries (backspace). - The result is displayed in an input field labeled “atm” for pressure. To solve this problem, apply the ideal gas law: \[ PV = nRT \] Where: - \( P \) = pressure (in atm) - \( V \) = volume (in liters, 5.00 L) - \( n \) = number of moles (7.95 moles) - \( R \) = ideal gas constant (0.0821 L·atm/mol·K) - \( T \) = temperature in Kelvin (39.3 °C + 273.15 = 312.45 K) Calculate the pressure \( P \) in atm using these values.
Transcribed Image Text:**Question 11 of 12** **Chemistry Problem:** *What is the pressure in atm of a 5.00 L tank with 7.95 moles of oxygen at 39.3 °C?* **Solution Interface:** - The diagram on the right shows a digital calculator interface. - The interface includes a numerical keypad with digits (1 to 9, 0), a decimal point (.), a button for exponentiation (x 10^), and operations for clearing (C) and deleting entries (backspace). - The result is displayed in an input field labeled “atm” for pressure. To solve this problem, apply the ideal gas law: \[ PV = nRT \] Where: - \( P \) = pressure (in atm) - \( V \) = volume (in liters, 5.00 L) - \( n \) = number of moles (7.95 moles) - \( R \) = ideal gas constant (0.0821 L·atm/mol·K) - \( T \) = temperature in Kelvin (39.3 °C + 273.15 = 312.45 K) Calculate the pressure \( P \) in atm using these values.
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