Solveeit Logo
Login

Question

Question: The compressibility of a gas in less than unity at STP. Therefore; A. \({V_m} > 22.4litre\) B. \...

The compressibility of a gas in less than unity at STP. Therefore;
A. Vm>22.4litre{V_m} > 22.4litre
B. Vm<22.4litre{V_m} < 22.4litre
C. Vm=22.4litre{V_m} = 22.4litre
D. Vm=44.8litre{V_m} = 44.8litre

Explanation

Solution

Hint- We will use the basic formula for the compressibility of gas and then substitute all the values into it. We will substitute the value of gas constant as 0.0821LatmK−10.0821Latm{K^{ - 1}}. The gas constant (also referred to as a molar, uniform or ideal gas constant) is referred to as the R symbol. This is equal to the Boltzmann constant, which is measured in units of energy per mole incremental temperature , i.e. the pressure-volume component, instead of energy per particle temperature change.
Formula used: Z=PVnRTZ = \dfrac{{PV}}{{nRT}}

Complete step-by-step answer:
As we know that the formula for compressibility of a gas is-
⇒Z=PVnRT\Rightarrow Z = \dfrac{{PV}}{{nRT}}
Where, Z stands for compressibility, P stands for the pressure of the gas, V stands for the molar volume of the gas and T stands for the temperature. R is the gas constant.
Since in question it is given that the compressibility of the gas is less than unity, we can say that-
⇒Z=PVnRT  ⇒PVnRT<1  ⇒PV<nRT  \Rightarrow Z = \dfrac{{PV}}{{nRT}} \\\ \\\ \Rightarrow \dfrac{{PV}}{{nRT}} < 1 \\\ \\\ \Rightarrow PV < nRT \\\
Now, substituting all the values into the above equation-
At STP,
P pressure of the gas = 1 atm
T temperature = 273K
R gas constant = 0.0821LatmK−10.0821Latm{K^{ - 1}}
⇒PV<nRT  ⇒1atm×V<1×0.0821×273  ⇒V<22.4litres  \Rightarrow PV < nRT \\\ \\\ \Rightarrow 1atm \times V < 1 \times 0.0821 \times 273 \\\ \\\ \Rightarrow V < 22.4litres \\\
Hence, option B is the correct option.

Note: Compressibility is a function of the relative volume increase in a fluid or object, often called the pressure (or mean stress) shift, in the field of thermodynamics and fluid mechanics. The variance of optimal gas efficiency appears to become particularly important near the critical point or in the case of high or low temperature (or, equivalently, the compressibility factor varies far from the unit). Throughout such situations, an objective finding may be derived from a simplified graph or alternative state equation that best fits the problem.