Chemistry, 06.11.2019 03:31 emmat1220

Asample 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.

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Answer from: zariyastanfield4452

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

Answer from: Quest

Igonna be honest i dunno that one srry

Answer from: Quest

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

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