Question

One mole of oxygen at STP is adiabatically compressed to 5 atm. Calculate the final
temperature. Also, calculate the work done on the gas. Take g = 1.4 and
R = 8.31 J mol−1 K−1.

Answer 1

QUESTION:

One mole of oxygen at STP (100, 273.15) $p_1 = 100 \, \text{kPa} \, T_1 = 273.15 \, \text{K}$ is adiabatically compressed to $p_2 = 5 \, \text{atm}$ (506.625 kPa). Calculate the final temperature $T_2$. Also, calculate the work done on the gas. Take $g = 1.4$ and $R = 8.31 \, \text{J mol}^{-1} \, \text{K}^{-1}$.

SOLUTION:

The mathematical equation for an ideal gas undergoing a reversible (i.e., no entropy generation) adiabatic process is

$$p_1 V_1^{\gamma} = \text{const}$$

where $P$ is pressure, $V$ is volume, and

$$\gamma = \frac{\alpha + 2}{\alpha} = 1.4 \text{ is the adiabatic index}$$

So:

$$p_1 V_1^{\gamma} = \text{const}$$ $$p_1 V_1 = n R T_1 \text{ (the ideal gas law)}$$

where $n$ is the amount of substance of gas (also known as number of moles), $T$ is the temperature of the gas and $R$ is the ideal, or universal, gas constant.

Hence:

$$V_1 = \frac{n R T_1}{p_1}$$ $$p_1 \left(\frac{n R T_1}{p_1}\right)^{\gamma} = \text{const}$$ $$p_1^{1 - \gamma} T_1^{\gamma} = \text{const}$$

Hence

$$p_1^{1 - \gamma} T_1^{\gamma} = p_2^{1 - \gamma} T_2^{\gamma}$$ $$T_2 = \left(\frac{p_1^{1 - \gamma} T_1^{\gamma}}{p_2^{1 - \gamma}}\right)^{\frac{1}{\gamma}}$$ $$T_2 = \left(\frac{(100 \cdot 10^3)^{1 - 1.4} \cdot 273.15^{1.4}}{(506.652 \cdot 10^3)^{1 - 1.4}}\right)^{\frac{1}{1.4}}$$ $$T_2 = 432.2 \, \text{K}$$

The work, done by the gas in adiabatic process is

$$W = - \frac{\alpha}{2} n R T_1 \left(\left(\frac{p_2}{p_1}\right)^{\frac{\gamma - 1}{\gamma}} - 1\right)$$

Since

$$\gamma = \frac{\alpha + 2}{\alpha}$$

We can find $\alpha$

$$\alpha \gamma = \alpha + 2$$ $$\alpha \gamma - \alpha = 2$$ $$\alpha (\gamma - 1) = 2$$ $$\alpha = \frac{2}{\gamma - 1} = 5$$

So, work done by the gas is

$$W = - \frac {5}{2} \cdot 1 \cdot 8.31 \cdot 273.15 \left( \left( \frac {506.625 \cdot 10^{3}}{100 \cdot 10^{3}} \right)^{\frac {1.4 - 1}{1.4}} - 1 \right)$$ $$W = -3.35 \text{ kJ}$$

And work done on the gas is $3.35 \text{ kJ}$

**ANSWER:**

$3.35 \text{ kJ}$