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Question: When \(110{\text{ g}}\) of manganese (Atomic mass \(55\)) dissolves in dilute \({\text{HN}}{{\text{O...

When 110 g110{\text{ g}} of manganese (Atomic mass 5555) dissolves in dilute HNO3{\text{HN}}{{\text{O}}_{\text{3}}}at 27oC{27^o}{\text{C}} under atmospheric pressure, the work done in the process is:
A.2494.2 J2494.2{\text{ J}}
B.−2494.2 J - 2494.2{\text{ J}}
C.−4988.4 J - 4988.4{\text{ J}}
D.4988.4 J4988.4{\text{ J}}

Explanation

Solution

Manganese on reaction with dilute nitric acid gets reduced by oxidizing agent nitric acid to give manganese nitrate and hydrogen gas. The gas produced changes the physical properties of the system and work is done. To answer this question, you must recall the ideal gas equation.
Formulae used:
PV=nRTPV = nRT
Where, PP is the pressure exerted by the gas on the walls of the container
VV is the total volume occupied by the gas or the volume of the container
nn is the number of moles of the gas present in the container
And, TT is the temperature at which the gas is present in the container

Complete step by step answer:
Using the ideal gas equation, at the given constant pressure and temperature conditions, we can write, W=PΔV=ΔngRTW = P\Delta V = \Delta {n_g}RT
The reaction occurring between manganese and nitric acid is given as:
Mn(s)+2HNO3(aq)→Mn(NO3)2(aq)+H2(g)Mn(s) + 2HN{O_3}(aq) \to Mn{\left( {N{O_3}} \right)_2}(aq) + {H_2}(g)
From the reaction, we can see that one mole of manganese gives one mole of hydrogen gas.
In the question, we have 110g110gof manganese. So the number of moles of manganese taken which will be equal to the number of moles of hydrogen produced are given by n=11055=2n = \dfrac{{110}}{{55}} = 2
So the work done during the process is given by W=ΔngRT=2×8.314×300W = \Delta {n_g}RT = 2 \times 8.314 \times 300
⇒W=−4988.4 J\Rightarrow W = - 4988.4{\text{ J}}

Hence, the correct answer is C.

Note:
Work done by a gas is given by Δ(PV)\Delta \left( {PV} \right). In the question, we do not have any information regarding the change in volume. The pressure and temperature are given as constant. The only change occurring is that in the number of gaseous substances present in the mixture.
The work done for an isobaric process is given as W=PΔVW = P\Delta V
Using the ideal gas equation, at constant pressure and temperature, we can write, W=PΔV=ΔnRTW = P\Delta V = \Delta nRT.