The homework question reads: "A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.

Explanation:

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Alleei

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Answer :  The work done for path A and path B is -685.3 J and -478.1 J  respectively.

Explanation :

To calculate the work done for path A :

First we have to calculate the moles of the gas.

where,

= initial pressure of gas  = 2.57 atm

= initial volume of gas  = 3.42 L

n = moles of gas  = ?

R = gas constant = 0.0821 atm.L/mol.K

T = temperature of gas  = 298 K

Now put all the given values in the above formula, we get:

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

where,

w = work done on the system = ?

n = number of moles of gas  = 0.359 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 298 K

= initial volume of gas  = 3.42 L

= final volume of gas  = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path A is, -685.3 J

To calculate the work done for path B :

The formula used for isothermally irreversible expansion is :

where,

w = work done

= external pressure = 1.19 atm

= initial volume of gas = 3.42 L

= final volume of gas = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path B is, -478.1 J

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :