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A sealed tank with a volume of 29.1 L contains 7.21 moles of propane gas at a temperature of 23.7 K. What is the pressure of the gas?

A) 0.299 atm
B) 0.687 atm
C) 2.432 atm
D) 4.645 atm

User Eric Liu
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8.3k points

1 Answer

6 votes

Final answer:

To find the pressure of the propane gas, the ideal gas law is used with the provided values, and after calculation, the pressure is determined to be 4.645 atm.

Step-by-step explanation:

To calculate the pressure of the propane gas in a sealed tank using the given values: volume (29.1 L), amount of substance (7.21 moles), and temperature (23.7 K), we use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

The value of R in liter-atmospheres per mole-Kelvin (L·atm/(mol·K)) is 0.0821. Rearranging the ideal gas law equation to solve for pressure (P), we get P = (nRT)/V. Plugging in the known values, we find P = (7.21 mol × 0.0821 L·atm/(mol·K) × 23.7 K) / 29.1 L, which simplifies to approximately P = 4.645 atm. Thus, the correct answer is D) 4.645 atm.

User Margo
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8.0k points
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