Question

At 27C , a gas is compressed suddenly such that its pressure becomes (1/8)th of its original pressure.Final temperature will be (r=5/3)
1)420k 2)300k 3)-142C 4)327C

Answer 1

Answer: (3). The final temperature will be 142 C.

Explanation:

Given that,

Temperature$T_{1}= 27+273=300 k$

$\gamma = \dfrac{5}{3}$

Pressure becomes (1/8)th of its original pressure.

We know that,

$T^{\gamma}P^{1-\gamma} = constant$

$T_{1}^{\gamma}P_{1}^{1-\gamma}=T_{2}^{\gamma}P_{2}^{1-\gamma}$

$\dfrac{P_{1}^{1-\gamma}}{P_{2}^{1-\gamma}}=\dfrac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

$(\dfrac{P_{1}}{P_{2}})^{1-\gamma}=(\dfrac{T_{2}}{T_{1}})^{\gamma}$

$(\dfrac{P\times8}{P})^{1-\dfrac{5}{3}}=(\dfrac{T_{2}}{300})^{\dfrac{5}{3}}$

$T_{2}= 300\times((8)^{-\dfrac{2}{3}})^{\dfrac{3}{5}}$

$T_{2}= 130.58\ K$

$T_{2}= 130.58-273 = -142.42\ C$

$T_{2}=-142\ C$

Hence, The final temperature will be -142 C.