A monoatomic gas at pressure P1 and volume V1 is compressed adiabatically to (1/8) textth of its original volume. What is the final pressure of the gas ?

Question

A monoatomic gas at pressure P1 and volume V1 is compressed adiabatically to (1/8) textth of its original volume. What is the final pressure of the gas ? (A) 64P1 (B) P1 (C) 16P1 (D) 32P1

Tardigrade Admin · Accepted Answer

Ideal gas equation, for an adiabatic process is P V&gamma;= constant or P1 V1&gamma;=P2 V2&gamma; For monoatomic gas &gamma;=(5/3) ∴ P1 V15 / 3=P2((V1/8))5 / 3 ⇒ P2=P1 ×(2)5=32 P1

Q. A monoatomic gas at pressure $P_{1}$ and volume $V_{1}$ is compressed adiabatically to $\frac{1}{8}^{th}$ of its original volume. What is the final pressure of the gas ?

$64 P_{1}$

$P_{1}$

$16 P_{1}$

$32 P_{1}$

Ideal gas equation, for an adiabatic process is
$P V^{γ} =$ constant or $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
For monoatomic gas $γ = \frac{5}{3}$
$∴ P_{1} V_{1}^{5/3} = P_{2} (\frac{V _{1}}{8})^{5/3}$
$\Rightarrow P_{2} = P_{1} \times (2)^{5} = 32 P_{1}$

Q. A monoatomic gas at pressure P1​ and volume V1​ is compressed adiabatically to 81​th of its original volume. What is the final pressure of the gas ?

64P1​

P1​

16P1​

32P1​