The universe has laws that the volume \([tex] V \text{ (m}^3) \[/tex]\) of a given mass of an ideal gas varies directly with its absolute temperature \([tex] T \text{ (K)} \[/tex]\) and inversely with its pressure [tex]\([tex] P \text{ (Pa)} \[/tex]\)[/tex].

Given: A certain gas at an absolute temperature of 275 K and a pressure of 105 N/m² has a volume of 0.0225 m³.

Find the formula for this condition.

The Ideal Gas Law is given by:
[tex]\[
PV = nRT
\][/tex]

Where:
- [tex]\( P \)[/tex] is the pressure
- [tex]\( V \)[/tex] is the volume
- [tex]\( n \)[/tex] is the number of moles
- [tex]\( R \)[/tex] is the ideal gas constant
- [tex]\( T \)[/tex] is the temperature in Kelvin



Answer :

The Ideal Gas Law is expressed with the formula:

[tex]\[ PV = nRT \][/tex]

where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the gas constant, and
- [tex]\( T \)[/tex] is the temperature of the gas.

Given Data:
- Volume ([tex]\( V \)[/tex]) = 0.0225 m³
- Temperature ([tex]\( T \)[/tex]) = 275 K
- Pressure ([tex]\( P \)[/tex]) = 105 N/m²
- Gas constant ([tex]\( R \)[/tex]) = 8.314 J/(mol·K)

To find the number of moles ([tex]\( n \)[/tex]), we need to rearrange the Ideal Gas Law to solve for [tex]\( n \)[/tex]:

[tex]\[ n = \frac{PV}{RT} \][/tex]

Step-by-Step Solution:
1. Identify known quantities:
[tex]\[ V = 0.0225 \text{ m}^3 \\ T = 275 \text{ K} \\ P = 105 \text{ N/m}^2 \\ R = 8.314 \text{ J/(mol·K)} \][/tex]

2. Substitute these values into the equation:
[tex]\[ n = \frac{105 \text{ N/m}^2 \times 0.0225 \text{ m}^3}{8.314 \text{ J/(mol·K)} \times 275 \text{ K}} \][/tex]

3. Calculate the numerator:
[tex]\[ 105 \text{ N/m}^2 \times 0.0225 \text{ m}^3 = 2.3625 \text{ N·m} = 2.3625 \text{ J} \quad (\text{since } 1 \text{ N·m} = 1 \text{ J}) \][/tex]

4. Calculate the denominator:
[tex]\[ 8.314 \text{ J/(mol·K)} \times 275 \text{ K} = 2286.35 \text{ J/mol} \][/tex]

5. Divide the numerator by the denominator:
[tex]\[ n = \frac{2.3625 \text{ J}}{2286.35 \text{ J/mol}} = 0.001033306361668161 \text{ mol} \][/tex]

So, the number of moles ([tex]\( n \)[/tex]) of the gas is approximately [tex]\( 0.001033 \)[/tex] mol.