Work done when 65.38g of zn is dissolved completely in hcl in an open

1.	Work done when 65.38g of zn is dissolved completely in hcl in an open beaker at 300k is
Answer» Answer:- work = -2496 J Solution:- The balanced equation for the reaction of zinc with HCl is: $Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$ GIVEN grams of Zn are CONVERTED to moles. Moles of hydrogen gas formed will ALSO be equal as there is 1:1 mol ratio between zinc and hydrogen. $65.38gZn(\frac{1molZn}{65.38gZn})(\frac{1molH_2}{1molZn})$ = $1molH_2$ From FIRST law of thermodynamics, $w=-p\Delta V$ .....(1) From an ideal gas law equation, PV = nRT So, $P\Delta V=\Delta nRT$ ---------(2) From equation 1 and 2: $w=-\Delta nRT$ From above calculation, $\Delta n$ is 1 mol. T = 300 K R = $0.0821\frac{atm.L}{mol.K}$ Let's plug in the values in the equation: $w=-1mol0.0821\frac{atm.L}{mol.K}300K$ w = -24.63 atm.L Since, 1 atm.L = 101.325 J So, $w=-24.63atm.L(\frac{101.325J}{atm.L})$ w = -2495.63 J or it is almost -2496 J The negative sign indicates the work done by the system which is also clear as the gas is formed means there is expansion in volume. So, the work done is -2496 J.