Question

one litre of nitrogen at 15 degree celsius and 10^6 Pa expands isothermally until its volume is doubled and then adiabatically until it is redoubled.find the final pressure of gas..

Answer 1

Answer on Question #45011 – Physics – Molecular Physics | Thermodynamics

Question.

One litre of nitrogen at 15 degree celsius and $10^{\wedge}6$ Pa expands isothermally until its volume is doubled and then adiabatically until it is redoubled. Find the final pressure of gas.

$$\begin{array}{l} V_0 = 1\,l \\ T_0 = 15{}^{\circ}\mathrm{C} \\ P_0 = 10^6\,Pa \\ V_1 = 2V_0 \\ V_2 = 2V_1 \\ P_2 = ? \end{array}$$

Solution.

The process $0 \to 1$ is isothermal. So, $T = \text{const}$.

We know the equation state of ideal gas:

$$PV = RT$$

For isothermal process we will receive:

$$PV = \text{const}$$

Therefore,

$$P_0 V_0 = P_1 V_1 = \text{const} \rightarrow P_1 = P_0 \frac{V_0}{V_1}$$

The process $1 \to 2$ is adiabatic. So, let consider the adiabatic equation:

$$PV^{\gamma} = \text{const}$$

Therefore,

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma} = \text{const} \rightarrow P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} = P_0 \frac{V_0}{V_1} \left(\frac{V_1}{V_2}\right)^{\gamma}$$

It remains to find the adiabatic index $\gamma$. From the data sheets it is known that the adiabatic index of $N_2$ at $15{}^{\circ}\mathrm{C}$ is equal to 1.404. So, $\gamma = 1.404$.

Calculate:

$$P _ {2} = 1 0 ^ {6} \cdot \frac {1}{2} \cdot \left(\frac {1}{2}\right) ^ {1. 4 0 4} = 1 0 ^ {6} \cdot 0. 5 \cdot 0. 3 7 8 = 0. 1 8 9 \cdot 1 0 ^ {6} P a = 0. 1 8 9 M P a$$

Answer.

$$P _ {2} = P _ {0} \frac {V _ {0}}{V _ {1}} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma} = 0. 1 8 9 \cdot 1 0 ^ {6} P a = 0. 1 8 9 M P a$$

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