[Win,odd Dhamnekar]

Van der waals' equation is [tex]\big( 1 bar+\frac{5.537 bar*l^2*mol}{V^2}\big)*(V-0.0305 l)=3099.4688722 J[/tex]

 We can express [itex]V= Xlitres[/itex] So, how to answer question 5)?

[Borek]

Molar volume of gaseous water is 0.030589 m³/mol using Charles' law for temperature- volume relationship.

You are not asked to use Charles' law for 1.

I pointed you in the right direction. You know the mass, you know the density, that's enough to calculate molar volume using a correct definition. If you prefer to ignore the advice and waste time - so be it.

[Win,odd Dhamnekar]

 I didn't know that the molar mass of liquid water and gaseous water are one and the same that is 18.02 g/mol.
 that's why i used charles' law.
 Any way, how to answer question 2) and 5) ?
  Regarding question 2), i don't have any idea.  But while answering question 5), I faced difficulty in inputing the equation into equation solver of HP 50g calculator. It said " Invalid object type". How to solve it manually for V in easy way?

[Win,odd Dhamnekar]

My attempt to answer question 5)
 Final Van der Waals' equation is [tex]100x_J+ \frac{553.7_{J·Mol}}{x}-3.05_J -\frac{16.88785_{J·Mol}}{x^2}=3099.4688722_J[/tex]

Where x= Volume in liters. [itex]1_{bar·liter}=100_J.[/itex] Now, how to find x?

[Win,odd Dhamnekar]

Answers 1) The molar volume of liquid water is [itex]0.0000188m^3/mol[/itex] viz. 0.0188 liters/mol.
   The molar volume of gaseous water is [itex]0.03053150574 m^3/mol[/itex] viz. 30.53150574 liters/mol.
2) The average distance between nearest-neighbor water molecules in the liquid phase is 315 pm and in the gaseous phase 3701 pm.

3) From ideal gas equation, the computed value of molar volume of gaseous water is [itex]0.030994688722 m^3/mol [/itex] The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.030994688722 m^3/mol}{0.03053150574 m^3/mol}=101.517[/itex]%

4) The molar volume of gaseous water at 1 bar and 372,78 K is [itex]0.0305338 m^3/mol[/itex] viz.30.5338 liters/mol. The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.0305338m^3/mol}{0.03053150574m^3/mol}=100.075144[/itex]%

5) The molar volume of gaseous water at 1 bar and 372.78 K from van der waals' equation is [itex]0.000140306968475 m^3/mol[/itex] viz.0.140306968475 liters/mol.

The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.000140306968475m^3/mol}{0.03053150574 m^3/mol}=0.459548[/itex]%

6)What are your comments on the results of questions 3,4 and 5? At this temperature, would you expect the accuracy to increase or decrease at lower pressure?