Q: 1 mm^{3} of a gas is compressed at 1 atmospheric pressure and temperature 27°C to 627°C. What is the final pressure under adiabatic condition (γ for the gas = 1.5)

(a) 27 × 10^{5} N⁄m^{2}

(b) 80 × 10^{5} N⁄m^{2}

(c) 36 × 10^{5} N⁄m^{2}

(d) 56 × 10^{5} N⁄m^{2}

**Click to See Answer : **

_{1}= 1 atm , T

_{1}= 27°C = 27 + 273 = 300 K

T_{2} = 627°C = 627 + 273 = 900 K, P_{2} = ?

$\large \frac{T_1^{\gamma}}{P_1^{\gamma -1}} = \frac{T_2^{\gamma}}{P_2^{\gamma -1}} $

$\large (\frac{P_2}{P_1})^{\gamma -1} = (\frac{T_2}{T_1})^{\gamma} $

$\large \frac{P_2}{P_1} = (\frac{T_2}{T_1})^\frac{\gamma}{\gamma -1} $

$\large \frac{P_2}{P_1} = (\frac{900}{300})^\frac{3/2}{\frac{3}{2} -1} $

$\large \frac{P_2}{P_1} = 3^3 = 27 $

P_{2} = 27 P_{1}

= 27 × 1 = 27 atm. = 27 × 10^{5} N⁄m^{2}