The volume of a given mass of gas changes with temperature and pressure. Therefore, to compare the volumes of different gases, they have to be taken or converted to same conditions of temperature and pressure using ideal gas equation. The most commonly used conditions are STP (Standard temperature and pressure) conditions.

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The volume of gas at STP is the volume occupied by gas at standard temperature and pressure. The standard value of temperature is 0^{o}C or 273.15 K and the standard value for pressure is 1 atm or 760 mm of Hg. Its value can be determined by using ideal gas equation.

Molar volume of gas at STP is the volume occupied by one mole of gas at standard temperature and pressure conditions.

It is the volume occupied by one mole of gas at a temperature of 0^{o}C or 273.15 Kelvin and pressure of 1 atm or 760 mm of Hg. Its value can be found out by applying ideal gas equation or equation of state for an ideal gas.

1 mole of gas molecule contains 6.023 x 10^{23} molecules. Therefore, it is the volume occupied by 6.023 x 10^{23} gas molecules at STP. Since 1 mole of each gas contains 6.023 x 1023 gas molecules, hence, molar volume of each gas at STP is same.

One mole gas at STP occupies 22.4 litres of volume.

When the temperature and pressure conditions in STP are taken as 0^{o}C and 1 atmosphere, the molar volume (volume occupied by 1 mole of gas) is 22.4 Litres or 22400ml.

Whereas when the temperature and pressure conditions at STP are taken as 0^{o}C and 1 bar, the molar volume (volume occupied by 1 mole of gas) is 22.7 Litres or 22700ml.

One mole of every ideal gas at standard temperature and pressure occupies same volume = 22.4 litres.

For example, molar volume of nitrogen gas at STP is 22.4 litres. It is also another definition for one mole of a gas.

The volume of gas at STP is found out by applying the ideal gas equation (PV = nRT) and using standard temperature and pressure condition.

At STP,

Let moles of gas = n moles

Pressure (P) = 1 atm

Temperature (T) = 0 0C = 273.15 K

R = 0.0821 litre $\frac{atm}{K.mol}$

Volume of gas = V

PV = nRT

V = $\frac{nRT}{P}$

**This can be understood by taking following example:**

**Example:** Find the volume of a 2.5 moles of oxygen gas at STP.

**Solution:**

At STP,

P = 1 atm

T = 273.15 K

n = 2.5 moles

R = 0.0821 litre $\frac{atm}{K.mol}$

Applying gas equation:

PV = nRT

V = $\frac{nRT}{P}$

V = $\frac{2.5 \times 0.0821 \times 273.15}{1}$ =**56.06 litres.**

To find molar volume of gas, ideal gas equation is applied for 1 mole and using the known values of temperature and pressure.

For example,

**Example:** Find the molar volume of Helium gas at a temperature of 25 ^{o}C and 1.5 atm pressure?

**Solution:**

Pressure (P) = 1.5 atm

Temperature (T) = 25^{o}C = 298.15 K

n = 1 mole

R = 0.0821 Litre atm/K. mol

Volume (V) =?

Using ideal gas equation

PV = $\frac{nRT}{P}$

V = $\frac{nRT}{P}$ = 1 $\times$ 0.0821 $\times$ $\frac{298.15}{1.5}$ =**16.32 Litres**

The volume of 1 mole of an ideal gas at STP can be determined by applying the ideal gas equation (PV = nRT) and using standard temperature and pressure condition.

At STP,

P = 1 atm

T = 0^{o}C = 273.15 K

n = 1 mole

R = $\frac{0.0821 \ litre \ atm}{K.mol}$

Applying ideal gas equation and substituting the values, we get

PV = nRT

V = $\frac{nRT}{P}$

V = (1mol) $\times$ $\frac{(0.0821 \ litre \ atm}{K.mol}$ $\times$ $\frac{(273.15 K)}{(1 \ atm)}$

V = 22.4 litres = 22400 ml

Hence, 22.4 litres or 22400 ml is the volume occupied by 1 mole of any ideal gas.

Molar volume of gas at STP is the volume occupied by one mole of gas at standard temperature and pressure conditions.

It is the volume occupied by one mole of gas at a temperature of 0

1 mole of gas molecule contains 6.023 x 10

One mole gas at STP occupies 22.4 litres of volume.

When the temperature and pressure conditions in STP are taken as 0

Whereas when the temperature and pressure conditions at STP are taken as 0

One mole of every ideal gas at standard temperature and pressure occupies same volume = 22.4 litres.

For example, molar volume of nitrogen gas at STP is 22.4 litres. It is also another definition for one mole of a gas.

The volume of gas at STP is found out by applying the ideal gas equation (PV = nRT) and using standard temperature and pressure condition.

At STP,

Let moles of gas = n moles

Pressure (P) = 1 atm

Temperature (T) = 0 0C = 273.15 K

R = 0.0821 litre $\frac{atm}{K.mol}$

Volume of gas = V

PV = nRT

V = $\frac{nRT}{P}$

At STP,

P = 1 atm

T = 273.15 K

n = 2.5 moles

R = 0.0821 litre $\frac{atm}{K.mol}$

Applying gas equation:

PV = nRT

V = $\frac{nRT}{P}$

V = $\frac{2.5 \times 0.0821 \times 273.15}{1}$ =

To find molar volume of gas, ideal gas equation is applied for 1 mole and using the known values of temperature and pressure.

For example,

Pressure (P) = 1.5 atm

Temperature (T) = 25

n = 1 mole

R = 0.0821 Litre atm/K. mol

Volume (V) =?

Using ideal gas equation

PV = $\frac{nRT}{P}$

V = $\frac{nRT}{P}$ = 1 $\times$ 0.0821 $\times$ $\frac{298.15}{1.5}$ =

The volume of 1 mole of an ideal gas at STP can be determined by applying the ideal gas equation (PV = nRT) and using standard temperature and pressure condition.

At STP,

P = 1 atm

T = 0

n = 1 mole

R = $\frac{0.0821 \ litre \ atm}{K.mol}$

Applying ideal gas equation and substituting the values, we get

PV = nRT

V = $\frac{nRT}{P}$

V = (1mol) $\times$ $\frac{(0.0821 \ litre \ atm}{K.mol}$ $\times$ $\frac{(273.15 K)}{(1 \ atm)}$

V = 22.4 litres = 22400 ml

Hence, 22.4 litres or 22400 ml is the volume occupied by 1 mole of any ideal gas.

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