Question

1.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#39781, 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 $g = 1.4$ and $R = 8.31 \, \text{J mol}^{-1} \, \text{K}^{-1}$.

Solution

The equation of adiabatic process for an ideal gas is

$$P V ^ {\gamma} = \text{const},$$

where $P$ is pressure, $V$ is volume, and $\gamma = 1.4$ - the adiabatic index of gas.

Also we know the state equation for an ideal gas:

$$P V = v R T,$$

where $v$ is the amount of substance of gas, $T$ is the temperature of the gas and $R$ is the universal gas constant.

So

$$V = \frac{v R T}{P} \quad \text{and} \quad P \left(\frac{v R T}{P}\right)^{\gamma} = \text{const} \rightarrow P^{1 - \gamma} \cdot T^{\gamma} = \text{const}.$$

Hence

$$P_{1}^{1 - \gamma} \cdot T_{1}^{\gamma} = P_{2}^{1 - \gamma} \cdot T_{2}^{\gamma} \rightarrow T_{2} = \left(\frac{P_{1}^{1 - \gamma} \cdot T_{1}^{\gamma}}{P_{2}^{1 - \gamma}}\right)^{\frac{1}{\gamma}}.$$

STP is $(P_{1} = 100 \, kPa, T_{1} = 273 \, K), P_{2} = 507 \, kPa.$

$$T_{2} = \left(\frac{(100 \cdot 10^{3})^{1 - 1.4} \cdot 273^{1.4}}{(507 \cdot 10^{3})^{1 - 1.4}}\right)^{\frac{1}{1.4}} = 434 \, K.$$

The work done on the gas in adiabatic process is

$$W = \frac{\alpha}{2} v R T_{1} \left(\left(\frac{P_{2}}{P_{1}}\right)^{\frac{\gamma - 1}{\gamma}} - 1\right),$$

where $\alpha = \frac{2}{\gamma - 1} = 5$

$$W = \frac{5}{2} \cdot 1 \cdot 8.31 \cdot 273 \left(\left(\frac{507 \cdot 10^{3}}{100 \cdot 10^{3}}\right)^{\frac{1.4 - 1}{1.4}} - 1\right) = 331 \, J.$$

Answer: 434 K; 331 J.