Question

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of 80.0 ∘c until the original pressure has tripled. calculate the amount of work done by gas.

Answer 1

The work done by a gas during an isothermal process is given by:
$W=nRT ln \frac{V_f}{V_i}$ (1)
where
n is the number of moles of the gas
R is the gas constant
T is the absolute temperature of the gas
$\frac{V_f}{V_i}$ is the ratio between the final volume and the initial volume of the gas

We need to calculate this ratio, and we can do it by using the gas pressure. In fact, for an isothermal process, Boyle's law states that the product between pressure and volume of the gas is constant:
$pV=k$
which can be rewritten as
$p_i V_i= p_f V_f$
which is equivalent to
$\frac{V_f}{V_i}= \frac{p_i}{p_f}$
The problem says that the pressure of the gas is tripled, therefore the ratio between final and initial volume is:
$\frac{V_f}{V_i} = \frac{p_i}{3 p_i} = \frac{1}{3}$

Now we can use eq.(1) to calculate the work done by the gas. The absolute temperature is
$T=80.0^{\circ}C+273 = 353 K$
The number of moles is n=2, therefore the work done is
$W=nRT ln \frac{V_f}{V_i}=(2 mol)(8.31 J/mol K) (353 K) \ln \frac{1}{3}= -6445 J$
And the work is negative, because it is done by the environment on the gas (the gas is compressed)