Question

the work done in an open vessel at 300 Kelvin when 56 gram iron react with dilute HCl is

Answer 1

We know that Molar mass of Fe = 56 and R = 2 cal^-1  K mol^-1

W = P(nRT/P)         n = 56/56 = 1

    = 1 * 2 * 300

    = 600 calorie.


Hope this helps!

Answer 2

The work done is 600 calories.

Given:

Temperature = 300 K

Mass of iron or Fe = 56 gm

Solution:

We know that work done can be calculated using W=PdV

Where,

PdV stands for Pressure Volume Work.

We also know that,

PV = nRT

Where,

PV = nRT is the ideal gas law.

Robert Boyle found that PV is a constant i.e., the pressure of a gas multiplied by the volume of a gas at any given temperature is a constant. Robert Boyle also found that since the temperature and number of moles does not change.  

So, PV = nRT

Where,

n is the number of moles  

T is the temperature

R is the gas constant and is equivalent to Boltzmann constant.

The value of R changes depending upon the units of pressure and volume used. In this case, we use the value of R = 2 $\mathrm{cal}^{-1} \mathrm{Kmol}^{-1}$

Now we know that 56 grams of iron is being heated at 300 K.

Thus,

$W=\frac{n R T}{\text {Mass of iron}}$

$W=\frac{56 \times 2 \times 300}{56}$

$W = 600 \ calories$