Question

The value of $PV$ for $5.6$ litre of an ideal gas is $\dfrac{1}{x}RT$ at STP.
Value of $x$ is $\_\_\_\_\_?$

Answer 1

Hint: For these question we have to know about the ideal gas equation the components in which that equation depends and the what is STP, the data are given $PV$ for $5.6$ litre what does it means and how we use ideal gas equation to solve this question.

Formula used:
Ideal gas equation,
$PV = nRT$
Where, $P$ is pressure, $V$ is volume, $T$ is temperature, $n$ is number of moles and $R$ is the gas constant.

Complete answer:
Let’s start with understanding the question, what the question is asking.
Given, the value of $PV$ for $5.6$ litre of an ideal gas is $\dfrac{1}{x}RT$ at STP.
What we understand by STP?
STP full form: STANDARD TEMPERATURE AND PRESSURE, STP is the condition at which the temperature is equal to $0^\circ $ Celsius and pressure is $1$ atmospheric.
As we know, from the moles concept,
$1$ Mole of an ideal gas occupies space of \[22.4\] litres at STP.
Here, in the question, $5.6$ Litre of an ideal gas is at STP.
So, by unitary method, we can find out the number of moles of the gases.
So, if $1$ moles contains \[22.4\] litres at STP.
$5.6$ Litre will contain $\dfrac{{5.6}}{{22.4}}$ mol
By solving by unitary method, the number of moles in the will be $0.25$ moles
But here, the number of moles in represented by $\dfrac{1}{x}$
Hence, we have to find the value of the $x$ ,
So, $\dfrac{1}{x} = 0.25$
From solving this, we find $x = 4$ .
Hence, the value of $PV$ for $5.6$ litre of an ideal gas is $\dfrac{1}{4}RT$ at STP.
Value of $x$ is $4$ .

Note:
The ideal gases are the gases which follow the Ideal gas equation. These gases do not exist in reality. This is a theoretical concept. The gases other than ideal gases are called real gases. The real gases at higher temperature and lower pressure can act as ideal gases.