Question

the volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be ( use R= 0.083 bar L K ^-1 mol^-1​

Answer 1

Answer:

=> 5.3701 L

Step-by-step explanation:

Given:-

mass of water vapour= 1.8g

Temperature=374°C

in kelvin, 374+273

=> 647k

Pressure of the container or water vapur= 1 bar.

(R to be user = 0.083 barLK^(-1)mol(^-1))

Using the ideal gas equation:-

=> PV=nRT

putting the values in equation...

Moles= g.m/m.m

=> n= 1.8/18 (Molar mass of H20= 1x2+16)

=> n= 1.8=18

=> n= 1/10

$(1) v = nrt \\ \\ v = \frac{1}{10} \times 0.083 \times 647 \\ \\ v = 0.083 \times 64.7 \\ \\ v = 5.3701$

So, the volume of water vapur is 5.3701 Litres.

Ans.

Answer 2

Given: 1.8 g of water vapour at 374° C and 1 bar pressure.

To find: The volume occupied by the water vapour.

Solution:

• According to the ideal gas equation,

        $PV = nRT$

• Here, P is the pressure, V is the volume of the water vapour, n is the number of moles present in 1.8 g, R is the gas constant and T is the temperature of the water vapour.
• The temperature (T) given in the question is 374° C and it must be converted to Kelvin as

        $374 + 273 = 647 K$

• If 18 g of water makes 1 mole, then 1.8 g would be 0.1 moles.

• On replacing the terms, the equation can be written as,

       $(1) (V) = (0.1) (0.083) (647)$

       $V = 5.37 L$

Therefore, the volume occupied by the water vapour is 5.37 L.