00:01
In this problem, we will take mass as 1 kg as everything is asked in terms of per kg.
00:10
So mass we will take as 1 kg.
00:14
Then the volume per unit mass of water or vmw is given as 1 .0 multiplied by 10 to the power minus 3 meter cube per kg.
00:31
So from here we can easily find out the density.
00:35
Of water which is just the reciprocal of the volume per minute mass so one by vmw which will be nothing but 10 to the power 3 kg per meter cube now since the mass is 1 kg from here we can find out the initial volume which will be vi which will be equal to mass divided by density of water so that will be one divided by 10 to the power 3 and the unit will be meter cube this will be 10 to the power minus 3 meter cube which will be 0 .001 meter cube this is v i or the initial volume now very easily from the first law of thermodynamics from the first law of thermodynamics from the first law of thermodynamics from the first law of thermodynamics we can say that del q the heat supplied is equal to del u the change in internal energy plus the work turn or w now we can calculate the value of del q we can calculate the value of del q because the formula will be the mass multiplied by the latent hit taken so in this case the mass is one kg and we will multiply that with the latent heat of vaporization which is given as 334 multiplied by 10 to the power 3 and the unit will be joules so this is nothing but 334 kilojoules.
02:29
This is del q.
02:32
Now we will have to find the final volume or vf.
02:37
To find the final volume, we can use the ideal gas equation.
02:43
So the ideal gas equation, which is p, the final volume, pv, will be equal to nrt, when is the number of moles and we will have to calculate the number of moles.
03:02
So for 1 kg, that is 1k .g that is 1 ,000 gram and divided by 18, which is the molar mass of water, this comes 55 .55 moles...