A sound wave passing through the air at NTP 0° C and 1 atm produces a pressure of 0.001 dyne cm -2 during a compression. The corresponding change in temperature (given γ=1.5 and assume gas to be ideal) is

Question

A sound wave passing through the air at NTP 0° C and 1 atm produces a pressure of 0.001 dyne cm -2 during a compression. The corresponding change in temperature (given &gamma;=1.5 and assume gas to be ideal) is (A) 8.97× 10- 4K (B) 9.87× 10- 6 K (C) 8.97× 10- 8 K (D) 9.87× 10- 4 K

Tardigrade Admin · Accepted Answer

T &gamma; P 1-&gamma;= constant Differentiating &gamma; T &gamma;-1 dT P1-&gamma;+ T &gamma; 1-&gamma; P -&gamma; dP =0 or, d T =(&gamma; -1 T /&gamma; P ) dP or, dT =(1.5-1/1.5) (273/76 × 13.6 × 981) × 0.001 ⇒ 8.97 × 10-8 K

Q. A sound wave passing through the air at NTP $0^{\circ} C$ and $1$ atm produces a pressure of $0.001$ dyne $c m^{- 2}$ during a compression. The corresponding change in temperature (given $γ = 1.5$ and assume gas to be ideal) is

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

$9.87 \times 1 0^{- 6} K$

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

$9.87 \times 1 0^{- 4} K$

$T^{γ} P^{1 - γ} =$ constant
Differentiating
$γ T^{γ - 1} d T P^{1 - γ} + T^{γ} 1 - γ P^{- γ} d P = 0$
or, $d T = \frac{γ - 1 ^{T}}{γ ^{P}} d P$
or, $d T = \frac{1.5 - 1}{1.5} \frac{273}{76 \times 13.6 \times 981} \times 0.001$
$\Rightarrow 8.97 \times 1 0^{- 8} K$

Q. A sound wave passing through the air at NTP 0∘C and 1 atm produces a pressure of 0.001 dyne cm−2 during a compression. The corresponding change in temperature (given γ=1.5 and assume gas to be ideal) is

8.97×10−4K

9.87×10−6K

8.97×10−8K

9.87×10−4K