Question

Calculate the temperature of $4.0$ moles of a gas occupying $5d{m^3}$ at $3.32bar$ ($R = 0.083bard{m^3}{K^{ - 1}}mo{l^{ - 1}}$)

Answer 1

The moles are the one of the main units in the chemistry. The moles of the molecule depend on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only. The number of equivalents of the reactant also depend on the moles of the molecule. The number of moles of the reactant and product are equal in the equilibrium reaction.
Formula used:
Moles are defined as the given mass of the molecule is divided by the molecular mass of the molecule.
${\text{moles}}{\text{ = }}\dfrac{{{\text{mass of the}}{\text{molecule}}}}{{{\text{molecular weight of the molecule}}}}$
The molecular weight of the molecule is dependent on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atom in the molecule.
$\text{Molecular Weight} = {\text{Number of the atoms}} \times {\text{Atomic weight of the atom}}$
The number of atoms of the molecules is equal to the number of moles of the molecules multiplied by Avogadro’s number.
$\text{The number of atoms} = {\text{number of moles}} \times 6.022 \times {10^{23}}$
The ideal gas equation depends on the pressure, temperature, number of moles, volume of the gas molecules in ideal condition.
The ideal gas equation is,
$PV = nRT$
Here, the pressure of the gas is P
The volume of the gas is V
The temperature of the gas in kelvin is T
Gas constant is R
The number of moles of the Gas molecules is n.

Complete answer:
The given data is
The pressure of the gas P is $3.32bar$
The volume of the gas V is $5d{m^3}$
Gas constant, $R = 0.083bard{m^3}{K^{ - 1}}mo{l^{ - 1}}$
The number of moles of the Gas molecules n is $4.0$ moles.
We calculate the temperature of the gas in kelvin T is given below,
The ideal gas equation is,
$PV = nRT$
We change the formula for our concern,
$T = \dfrac{{PV}}{{nR}}$
We substitute the known values from the given data.
$T = \dfrac{{3.32bar \times 5d{m^3}}}{{0.083bard{m^3}{K^{ - 1}}mo{l^{ - 1}} \times 4.0mol}}$
On simplification we get,
$T = 50K$
The temperature of $4.0$ moles of a gas occupying $5d{m^3}$ at $3.32bar$ and$R = 0.083bard{m^3}{K^{ - 1}}mo{l^{ - 1}}$ is $50K$.
According to the above discussion, we conclude the temperature of $4.0$ moles of a gas occupying $5d{m^3}$ at $3.32bar$ and $R = 0.083bard{m^3}{K^{ - 1}}mo{l^{ - 1}}$ is $50K$.

Note:
As we know that all the gases are not ideal in nature. Depending on the condition of the gas it behaves as ideal gas. There are three units for measuring the temperature. There are degrees Celsius, kelvin and Fahrenheit. Charles is one of the important laws in the gaseous state. Charles' law was proposed by J.A.C. Charles. This law is used to study the relationship volume of a gas and its temperature. In Charles' law the mass of the system and pressure of the system is constant. The Combination of Charles law, Boyle's law and Avogadro’s hypothesis is known as the ideal gas equation.