A sound wave passing through an ideal gas at NTP produces a pressure change of 0.001 dyne/ cm 2 during adiabatic compression. The corresponding change in temperature (γ=1.5 for the gas and atmospheric pressure is 1.013 × 106 dyne/cm 2 ) is

Question

A sound wave passing through an ideal gas at NTP produces a pressure change of 0.001 dyne/ cm 2 during adiabatic compression. The corresponding change in temperature (&gamma;=1.5 for the gas and atmospheric pressure is 1.013 × 106 dyne/cm 2 ) is (A) 8.97 × 10-4 K (B) 8.97 × 10-6 K (C) 8.97 × 10-8 K (D) 8.97 × 10-9 K

Tardigrade Admin · Accepted Answer

T&gamma; p1-&gamma;= constant or T&gamma;=p&gamma;-1 T=p((&gamma;-1/&gamma;)) ∴ (Δ T/T) =(&gamma;-1/&gamma;) × (Δ p/p) (Δ T/T) =((1.5-1/1.5)) × (0.001/1.013 × 106) Δ T=8.98 × 10-8 K

Q. A sound wave passing through an ideal gas at NTP produces a pressure change of $0.001$ dyne/ $c m^{2}$ during adiabatic compression. The corresponding change in temperature $(γ = 1.5$ for the gas and atmospheric pressure is $1.013 \times 1 0^{6}$ dyne/cm $^{2}$ ) is

$8.97 \times 1 0^{- 4} K$

$8.97 \times 1 0^{- 6} K$

$8.97 \times 1 0^{- 8} K$

$8.97 \times 1 0^{- 9} K$

$T^{γ} p^{1 - γ} =$ constant
or $T^{γ} = p^{γ - 1}$
$T = p^{(\frac{γ - 1}{γ})}$
$∴ \frac{Δ T}{T} = \frac{γ - 1}{γ} \times \frac{Δ p}{p}$
$\frac{Δ T}{T} = (\frac{1.5 - 1}{1.5}) \times \frac{0.001}{1.013 \times 1 0 ^{6}}$
$Δ T = 8.98 \times 1 0^{- 8} K$

Q. A sound wave passing through an ideal gas at NTP produces a pressure change of 0.001 dyne/ cm2 during adiabatic compression. The corresponding change in temperature (γ=1.5 for the gas and atmospheric pressure is 1.013×106 dyne/cm 2 ) is

8.97×10−4K

8.97×10−6K

8.97×10−8K

8.97×10−9K