Question

Calculate the maximum work done when pressure on 10g of hydrogen is reduced from 20 to 1 atm at a constant tempture of 273K. The gas behaves ideally.

Answer 1

Answer : The maximum work done is, $-34003.52J$

Solution :

This is the case of isothermal reversible expansion of gas.

Formula used :

$w=-2.303nRT\log (\frac{P_1}{P_2})\\\\w=-2.303\times \frac{m}{M}\times RT\log (\frac{P_1}{P_2})$

where,

w = work done by the gas

n = number of moles of gas

R = gas constant = 8.314 J/mole K

T = temperature of gas = 273 K

m = mass of hydrogen gas = 10 g

M = molar mass of hydrogen gas, $(H_2)$ = 2 g/mole

$P_1$ = initial pressure of gas = 20 atm

$P_2$ = final pressure of gas = 1 atm

Now put all the given values in the above formula, we get the work done.

$w=-2.303\times \frac{10g}{2g/mole}\times 8.314J/moleK\times 273K\times \log (\frac{20atm}{1atm})$

$w=-34003.52J$

Therefore, the maximum work done is, $-34003.52J$