Question

Find the initial temperature in K of a gas if the final temperature is 500 degrees Celsius and is expanded adiabatically by a pressure ration of 11:1 the adiabatic index is 1.38

Answer 1

Answer on Question #45009 – Physics – Molecular Physics | Thermodynamics

Question.

Find the initial temperature in K of a gas if the final temperature is 500 degrees Celsius and is expanded adiabatically by a pressure ration of 11:1 the adiabatic index is 1.38.

$$T _ {2} = 500{}^{\circ} \mathrm{C} = 773\ \mathrm{K}$$ $$\frac {P _ {1}}{P _ {2}} = \frac {11}{1} = 11$$ $$\gamma = 1.38$$ $$T _ {1} = ?$$

Solution.

The adiabatic equation:

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

But from the equation of ideal gas we know

$$P V = R T \rightarrow \frac {P V}{T} = R = \text{const} \rightarrow V = \frac {\text{const}}{P} T$$

Therefore,

$$P ^ {1 - \gamma} T ^ {\gamma} = \text{const}$$

In our case,

$$P _ {1} ^ {1 - \gamma} T _ {1} ^ {\gamma} = P _ {2} ^ {1 - \gamma} T _ {2} ^ {\gamma} = \text{const} \rightarrow \left(\frac {T _ {1}}{T _ {2}}\right) ^ {\gamma} = \left(\frac {P _ {2}}{P _ {1}}\right) ^ {1 - \gamma}$$

So,

$$T _ {1} = T _ {2} \left(\frac {P _ {1}}{P _ {2}}\right) ^ {\frac {\gamma - 1}{\gamma}}$$

Calculate:

$$T _ {1} = 773 \cdot 11^{0.275} = 773 \cdot 1.934 = 1495\ \mathrm{K}$$

Answer.

$$T _ {1} = T _ {2} \left(\frac {P _ {1}}{P _ {2}}\right) ^ {\frac {\gamma - 1}{\gamma}} = 1495\ \mathrm{K}$$

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