Solveeit Logo

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 PVPV for 5.65.6 litre of an ideal gas is 1xRT\dfrac{1}{x}RT at STP.
Value of xx is \\_\\_\\_\\_\\_?

Explanation

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 PVPV for 5.65.6 litre what does it means and how we use ideal gas equation to solve this question.

Formula used:
Ideal gas equation,
PV=nRTPV = nRT
Where, PP is pressure, VV is volume, TT is temperature, nn is number of moles and RR is the gas constant.

Complete answer:
Let’s start with understanding the question, what the question is asking.
Given, the value of PVPV for 5.65.6 litre of an ideal gas is 1xRT\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 00^\circ Celsius and pressure is 11 atmospheric.
As we know, from the moles concept,
11 Mole of an ideal gas occupies space of 22.422.4 litres at STP.
Here, in the question, 5.65.6 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 11 moles contains 22.422.4 litres at STP.
5.65.6 Litre will contain 5.622.4\dfrac{{5.6}}{{22.4}} mol
By solving by unitary method, the number of moles in the will be 0.250.25 moles
But here, the number of moles in represented by 1x\dfrac{1}{x}
Hence, we have to find the value of the xx ,
So, 1x=0.25\dfrac{1}{x} = 0.25
From solving this, we find x=4x = 4 .
Hence, the value of PVPV for 5.65.6 litre of an ideal gas is 14RT\dfrac{1}{4}RT at STP.
Value of xx is 44 .

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.