
   V = \frac{nRT}{p} = \frac{0,016625 \, \text{mol} \times 8,314 \, \text{J/(mol·K)} \times 273,15 \, \text{K}}{101 325 \, \text{Pa}}


The volume \( V \) is approximately \( 0.000373 \, \text{m}^3 \) or \( 373 \, \text{cm}^3 \), based on the provided values.

Would you like more details or have any other questions?

Here are 5 related questions to consider:
1. How does temperature affect the volume in the ideal gas law?
2. What is the significance of the gas constant \( R \) in different units?
3. How would the volume change if the pressure was halved?
4. Can this equation be applied to real gases under all conditions?
5. How do you convert volume in \( \text{m}^3 \) to liters?

**Tip:** Remember that standard conditions for gases (STP) are 273.15 K and 1 atm pressure (101,325 Pa).

V = \frac{nRT}{p} = \frac{0,016625 \text{mol} \times 8,314 \text{J/(mol·K)} \times 273,15 \text{K}}{101 325 \text{Pa}}

This example demonstrates how to calculate the volume of a gas using the ideal gas law, with given values for moles, temperature, and pressure. Learn the step-by-step process for finding the volume of the gas, including the application of the ideal gas law formula V = nRT/p. Suitable for high school students studying chemistry or physics.

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