Question

There is standard value of temperature and pressure at which the molar volume of gas is 22.4 L. The correct values are:
A.\[\;273{\text{ }}K,{\text{ }}1{\text{ }}atm\]
B.\[300{\text{ }}K,{\text{ }}760{\text{ }}mm\]
C.\[{25^0}C,{\text{ }}760{\text{ }}mm\]
D.\[373{\text{ }}K,{\text{ }}1{\text{ }}atm\]

Answer 1

Hint: The values are defined clearly by STP i.e. standard temperature pressure and also the temperature is at which water freezes and the pressure is normal atmospheric pressure that we consider daily. These are the STP parameters and these will be the required values in question.

Step by step explanation:In the given question, molar volume of a gas is given and this volume is occurring at standard temperature and pressure which are explained by IUPAC as STP i.e. Standard temperature and pressure.
So, there are few conditions in chemistry which are used in different experiments and calculations and they are STP i.e. standard temperature pressure and other one is NTP i.e. Normal temperature pressure and the parameters of temperature and pressure are already defined for these and by using these parameters we can easily calculate the value of volume of gas i.e. molar volume.
So, these standard conditions are as follows: -
Temperature = \[{0^0}C\]and we know that it is equivalent to 273 Kelvin (simple conversion of temperature from Celsius to kelvin)
Pressure = 1 atm
So, the correct answer is A i.e. 273 k and 1 atm.

Note:We can also verify this data and find out the molar volume of gas by using the ideal gas equation which is \[PV{\text{ }} = {\text{ }}nRT\]in which P = pressure, V = volume, n = number of moles, R = universal gas constant and T = temperature
Using the standard values i.e. STP and for 1 mole of gas
$V = \dfrac{{nRT}}{P} = \dfrac{{1 \times 0.082Latmmo{l^{ - 1}}{K^{ - 1}} \times 273K}}{{1atm}}$= 22.4 L
So, we have also proved that at STP, the value of molar volume of a gas is 22.4 L