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Question: What is the pressure in the atmosphere of 0.246 gram of hydrogen gas occupying a volume of 0.0500 li...

What is the pressure in the atmosphere of 0.246 gram of hydrogen gas occupying a volume of 0.0500 liters at 20C{{20}^{\circ }}C?

Explanation

Solution

To solve the given question we can use the ideal gas equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
While finding the number of moles, take the molecular mass of hydrogen gas as 2.016 g/ mol because in gaseous form the formula of hydrogen is H2{{H}_{2}}.

Complete answer:
We have to find the pressure of the gas in the atmosphere and we are given the values of volume, temperature, and given mass. So to solve the given question we can use the ideal gas equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, we have to find the number of moles, as the given mass of hydrogen gas is 0.246 grams and the molecular of hydrogen will be:
Molecular mass = 1.008 x 2 = 2.016 g/ mol
This is due to the fact that in gaseous form the formula of hydrogen is H2{{H}_{2}}.
The number of moles will be:
Moles=0.2462.016=0.122Moles=\dfrac{0.246}{2.016}=0.122
The temperature will be = 21 + 273 = 294 K
Volume given is 0.0500 Liters
Gas constant will be 0.08205 L atm K1 mol1\text{L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}
Putting all the values in the formula, we get:
P x 0.0500 = 0.122 x 0.8205 x 294P\text{ x 0}\text{.0500 = 0}\text{.122 x 0}\text{.8205 x 294}
P = 0.122 x 0.8205 x 2940.0500=58.9P\text{ =}\dfrac{\text{ 0}\text{.122 x 0}\text{.8205 x 294}}{\text{0}\text{.0500}}=58.9
So, the pressure of the gas will be 58.9 atm.

Note:
Whenever you are using the ideal gas formula then take the temperature in Kelvin only and the number of moles will be calculated by dividing the given mass by the molecular or atomic mass of the given compound.