Question

a quantity 4.3 liter of an ideal gas at pressure 2atm is compressed adiabatically to volume 1 liter find
1-the final pressure
2-work done the gas
take gama constant is 1.4

Answer 1

Answer on Question #43115 – Physics – Molecular Physics | Thermodynamics

Question.

a quantity 4.3 liter of an ideal gas at pressure 2 atm is compressed adiabatically to volume 1 liter

find

1-the final pressure

2-work done the gas

take gamma constant is 1.4

Given:

$V_{0} = 4.3\ l$ is the initial volume

$P_{0} = 2\ atm$ is the initial pressure

$V = 1\ l$ is the final volume

$\gamma = 1.4$ is the adiabatic constant

Find:

1) $P = ?$ the final pressure

2) $A = ?$ the work done the gas

Solution.

1) The adiabatic process equation:

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

So,

$$P _ {0} V _ {0} ^ {\gamma} = P V ^ {\gamma} \rightarrow P = P _ {0} \left(\frac {V _ {0}}{V}\right) ^ {\gamma}$$

Calculate:

$$P = 2 \cdot \left(\frac {4 . 3}{1}\right) ^ {1. 4} = 2 \cdot 7. 7 = 1 5. 4\ atm$$

2) By definition work done is:

$$A = \int_{V _ {0}}^{V} P d V$$

In our case,

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

But, $\text{const} = P_0 V_0^\gamma$. Therefore,

$$P = \frac{P_0 V_0^\gamma}{V^\gamma}$$

Calculate the integral to define the work done:

$$\begin{aligned} A &= \int_{V_0}^V P \, dV = P_0 V_0^\gamma \int_{V_0}^V \frac{dV}{V^\gamma} = P_0 V_0^\gamma \frac{1}{1 - \gamma} \frac{1}{V^{\gamma-1}} \Big|_{V_0}^V = \frac{P_0 V_0^\gamma}{1 - \gamma} \left( \frac{1}{V^{\gamma-1}} - \frac{1}{V_0^{\gamma-1}} \right) = \\ &= \frac{P_0 V_0}{1 - \gamma} \left( \left( \frac{V_0}{V} \right)^{\gamma-1} - 1 \right) \end{aligned}$$

Calculate:

$$\begin{aligned} A &= \frac{2 \cdot 4.3}{1 - 1.4} \left( \left( \frac{4.3}{1} \right)^{1.4 - 1} - 1 \right) = - \frac{8.6}{0.4} (7.7 - 1) = -21.5 \cdot 6.7 = -144 \, \text{atm} \cdot l \\ &\quad 1 \, \text{atm} = 101300 \, \text{Pa}; \, 1 \, l = 10^{-3} \, \text{m}^3 \end{aligned}$$ $$A = -144 \, \text{atm} \cdot l = -144 \cdot 101300 \cdot 10^{-3} = 14587 \, \text{Pa} \cdot \text{m}^3 = 14587 \, J = 14.587 \, kJ$$

Answer.

1)

$$P = P_0 \left( \frac{V_0}{V} \right)^\gamma = 15.4 \, \text{atm}$$

2)

$$A = \frac{P_0 V_0}{1 - \gamma} \left( \left( \frac{V_0}{V} \right)^{\gamma-1} - 1 \right) = 14587 \, J = 14.587 \, kJ$$

http://www.AssignmentExpert.com/