Question 897: Thermodynamics & KTG

Question

A monoatomic gas is suddenly compressed to $$(1 / 8)^{	ext {th }}$$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is $$(\gamma=5 / 3)$$

Accepted Answer

For a monoatomic gas undergoing adiabatic compression, we use the adiabatic relation:
$$ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} $$
Given that the gas is compressed to $$(1/8)$$ of its initial volume, we have:
$$ V_2 = \frac{V_1}{8} $$
We need to find the ratio of the final pressure $$(P_2)$$ to the initial pressure $$(P_1)$$. Substituting the volumes into the adiabatic relation, we get:
$$ P_1 V_1^{\gamma} = P_2 \left(\frac{V_1}{8}\right)^{\gamma} $$
Let's solve for $$(P_2 / P_1)$$:
$$ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} $$
Since $$(V_2 = V_1 / 8)$$, we have:
$$ \frac{V_1}{V_2} = 8 $$
Therefore:
$$ P_2 = P_1 \cdot 8^{\gamma} $$
Given $$(\gamma = 5/3)$$, we need to calculate $$(8^{\gamma})$$:
$$ 8^{5/3} = (2^3)^{5/3} = 2^5 = 32 $$
Thus, the ratio of the final pressure to the initial pressure is:
$$ \frac{P_2}{P_1} = 32 $$
The correct answer is Option A: 32.

Answer

32

Answer

8

Answer

$$\frac{40}{3}$$

Answer

$$\frac{24}{5}$$

Thermodynamics & KTG

A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }} of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is (\gamma=5 / 3)

Choose the correct answer: