Question

A certain mass of a gas at 273 K temperature and one atmospheric pressure is expanded
to 3 times its original volume under adiabatic conditions. Calculate the resulting
temperature and pressure. (Take the value of γ= 1.4)

Answer 1

Answer on Question #50649 – Physics – Molecular Physics | Thermodynamics

Question.

A certain mass of a gas at 273 K temperature and one atmospheric pressure is expanded to 3 times its original volume under adiabatic conditions. Calculate the resulting temperature and pressure. (Take the value of $\gamma = 1.4$ )

Given:

$T_{1} = 273K$

$P_{1} = 1$ atm

$V_{2} = 3V_{1}$

$\gamma = 1.4$

Find:

$T_{2} = ?$

$P_{2} = ?$

Solution.

Adiabatic process is described by the following equation:

$$P V ^ {\gamma} = c o n s t$$

Therefore,

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

From other hand, the adiabatic equation is:

$$T V ^ {\gamma - 1} = c o n s t$$

Therefore,

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

Calculate:

$$P _ {2} = P _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma} = 1 a t m \cdot \left(\frac {1}{3}\right) ^ {1. 4} = 1 a t m \cdot 0. 2 1 5 = 0. 2 1 5 a t m$$ $$T _ {2} = T _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma - 1} = 2 7 3 \cdot \left(\frac {1}{3}\right) ^ {0. 4} = 2 7 3 \cdot 0. 6 4 4 = 1 7 5. 8 K$$

Answer.

$$P _ {2} = P _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma} = 0.215 \text{ atm}$$ $$T _ {2} = T _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma - 1} = 175.8 \text{ K}$$

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