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

Answer on Question #40734, Physics, Molecular Physics

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 $\gamma = 1.4$ and $R = 8.31 \, \text{J mol}^{-1} \, \text{K}^{-1}$.

Solution:

**Given:**

Standard temperature and pressure (informally abbreviated as STP) is

$$T_1 = 273.15 \, \text{K}$$ $$p_1 = 100 \, \text{kPa}$$ $$p_2 = 5 \, \text{atm} = 5 \cdot 101325 \, \text{Pa} = 506.625 \, \text{kPa}$$ $$T_1 = ?$$

An adiabatic process is one in which no heat is gained or lost by the system.

The adiabatic condition

$$p V^{\gamma} = \text{const}$$

where $p$ is pressure, $V$ is volume, and $\gamma = 1.4$ is the adiabatic index.

An ideal gas can be characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them may be deduced from kinetic theory and is called the ideal gas law:

$$p V = n R T$$

where $n =$ number of moles, $R =$ universal gas constant $= 8.31 \, \text{J/mol K}$.

**Hence:**

$$V = \frac{n R T}{p}$$ $$p \left(\frac{n R T}{p}\right)^{\gamma} = \text{const}$$ $$p^{1 - \gamma} T^{\gamma} = \text{const}$$

**Hence**

$$p_1^{1 - \gamma} T_1^{\gamma} = p_2^{1 - \gamma} T_2^{\gamma}$$ $$T_2 = T_1 \left(\frac{p_1^{1 - \gamma}}{p_2^{1 - \gamma}}\right)^{\frac{1}{\gamma}}$$ $$T_2 = 273.15 \cdot \left(\frac{100^{1 - 1.4}}{506.625^{1 - 1.4}}\right)^{\frac{1}{1.4}} = 434.25 \, \text{K}$$

The work, done by the gas in adiabatic process is

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

where $\alpha$ is the number of degrees of freedom divided by two.

$\alpha$ is 5/2 for diatomic gas such as oxygen.

Thus,

$$W _ {b} = - \frac {5}{2} \cdot 1 \cdot 8.31 \cdot 273.15 \cdot \left(\left(\frac {506.625}{100}\right) ^ {\frac {1.4 - 1}{1.4}} - 1\right) = -3346.86 \text{ J}$$

And work done on the gas

$$W = - W _ {b} = 3.35 \text{ kJ}$$

Answer. $T_{2} = 434.25 \text{ K}, W = 3.35 \text{ kJ}$.