Question

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 fi nal volume is 7.39 L. Calculate the work for each path.

Answer 1

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.

$PV=nRT$

where,

$P_1$ = initial pressure of gas  = 2.57 atm

$V_1$ = 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:

$PV=nRT$

$(2.57atm)\times (3.42L)=n\times (0.0821atm.L/mol.K)\times (298K)$

$n=0.359mole$

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

As per first law of thermodynamic,

$\Delta U=q+w$

where,

$\Delta U$ = 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.

$\Delta U=0$

$q=-w$

The expression used for work done will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

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

$V_1$ = initial volume of gas  = 3.42 L

$V_2$ = final volume of gas  = 7.39 L

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

$w=-0.359mole\times 8.314J/moleK\times 298K\times \ln (\frac{7.39L}{3.42L})$

$w=-685.3J$

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 :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure = 1.19 atm

$V_1$ = initial volume of gas = 3.42 L

$V_2$ = final volume of gas = 7.39 L

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

$w=-p_{ext}(V_2-V_1)$

$w=-(1.19atm)\times (7.39-3.42)L$

$w=-4.72L.atm=-4.72\times 101.3J=-478.1J$

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