Respuesta :
The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.
The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.
The scalar quantity that tells about the space occupied by the ideal hypothetical gas is called volume. The volume required by the 3.40 mol of gas is 3.85 L.
What is the ideal gas equation?
An ideal gas equation is the hypothetical gas relation of the pressure, volume, moles and the temperature of the gas of the system. It is given by,
[tex]\rm PV = nRT[/tex]
Where,
Pressure (P) = 22.2 atm
Volume = V
Temperature (T) = 306.45 K
Gas constant (R) = 0.0821
Moles (n) = 3.40 mol
Substituting values in the above equation:
[tex]\begin{aligned}\rm V &=\rm \dfrac{nRT}{P}\\\\&= \dfrac{3.4 \times 0.0821 \times 306.45}{22.2}\\\\&= 3.85 \;\rm L\end{aligned}[/tex]
Therefore, 3.85 L is the volume of the 3.40 mol of gas.
Learn more about the ideal gas equation here:
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