Question
Question: The value of \(PV\) for \(5.6\) litre of an ideal gas is \(\dfrac{1}{x}RT\) at STP. Value of \(x\)...
The value of PV for 5.6 litre of an ideal gas is x1RT at STP.
Value of x is \\_\\_\\_\\_\\_?
Solution
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 x1RT 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∘ 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 22.45.6 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 x1
Hence, we have to find the value of the x ,
So, x1=0.25
From solving this, we find x=4 .
Hence, the value of PV for 5.6 litre of an ideal gas is 41RT 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.