Question

A scientist observed that a gas expanded from a volume of 0.200 L to a volume of 0.876 L. (a) What is the amount of work (in joules) performed in this process if the gas expanded at a constant pressure of 0.389 atm? w = J (b) If the temperature of the gas did not change during the expansion, calculate the change in internal energy of the gas, as well as the change in heat due to the expansion. ΔE = J q = J

Answer 1

Answer :

(a) The amount of work performed in this process is -26.6 J

(b) The value of change in internal energy is 0 J and change in heat is 26.6 J

Explanation :

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 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 = 0.389 atm

$V_1$ = initial volume of gas = 0.200 L

$V_2$ = final volume of gas = 0.876 L

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

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

$w=-(0.389atm)\times (0.876-0.200)L$

$w=-0.263L.atm=-0.263\times 101.3J=-26.6J$

The work done by the system on the surroundings are, -26.6 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the amount of work performed in this process is -26.6 Joules.

q = - w = -(-26.6 J) = 26.6 J

Net heat transfer is, 26.6 J