Question

Two moles of a perfect gas occupy a volume 5
of 0.065 m3 and exert a pressure of
3.0 x 105 N m - 2. It is compressed
isobarically to 0.050 m3. Determine the
work done by the gas and the fall in its
temperature. Take R=8.3 JK - I mol -1.

Answer 1

Two moles of a perfect gas occupy a volume 5 of $0.065\,\mathrm{m}^3$ and exert a pressure of $3.0 \times 10^{5}\,\mathrm{N}$ $\mathrm{m}^{-2}$. It is compressed isobarically to $0.050\,\mathrm{m}^3$. Determine the work done by the gas and the fall in its temperature. Take $R = 8.3\,\mathrm{JK} \cdot \mathrm{L}\,\mathrm{mol}^{-1}$.

Solution

The work done by the gas that isobarically compressed:

$$W = P \Delta V = P (V_2 - V_1),$$

where $P$ – pressure, $V_1$ – initial volume of a perfect gas, $V_2$ – final volume of a perfect gas.

So

$$W = 3.0 \times 10^5 \frac{\mathrm{N}}{\mathrm{m}^2} (0.050\,\mathrm{m}^3 - 0.065\,\mathrm{m}^3) = -4.5 \times 10^3 J.$$

A sign “-” means that the work done on the gas.

Let’s write state equation of perfect gas for initial and final states:

$$PV_1 = vRT_1 \quad \text{and} \quad PV_2 = vRT_2.$$

The fall in the temperature of perfect gas:

$$T_1 - T_2 = \frac{PV_1 - PV_2}{vR} = \frac{3.0 \times 10^5 \frac{\mathrm{N}}{\mathrm{m}^2} (0.065\,\mathrm{m}^3 - 0.050\,\mathrm{m}^3)}{2\,\mathrm{mol} \cdot 8.3 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}} = 271\,\mathrm{K}.$$

Answer: $-4.5 \cdot 10^3 J$; $271\,\mathrm{K}$.