A fixed amount of dry air temperature of 27-o-C is compressed to cfrac-1-9-th of original volume- Its final temperature is left - gamma -1-5 right -627-o-C600-o-C527-o-C158-o-C

Question

Toppr Admin · Accepted Answer

In an adiabatic process-TV-x3B3-x2212-1-constantT1V-x3B3-x2212-11-T2V-x3B3-x2212-12T2-T1-xD7-V-x3B3-x2212-11V-x3B3-x2212-12-300-xD7-11-9-1-5-x2212-1-300-3-900K-627oCOption A-

A fixed amount of dry air at temperature of $27^{o} C$ is compressed to $\frac{1}{9} t h$ of original volume. Its final temperature is $(γ = 1.5)$
$627^{o} C$
$600^{o} C$
$527^{o} C$
$158^{o} C$

In an adiabatic process,
$T V^{γ - 1} =$ constant
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$T_{2} = T_{1} \times \frac{V_{1}^{γ - 1}}{V_{2}^{γ - 1}} = 300 \times {(\frac{1}{1 / 9})}^{1.5 - 1} = 300 (3) = 900 K = 627^{o} C$
Option A.

A fixed amount of dry air at temperature of 27oC is compressed to 19th of original volume. Its final temperature is (γ=1.5)627oC600oC527oC158oC

In an adiabatic process,TVγ−1=constantT1Vγ−11=T2Vγ−12T2=T1×Vγ−11Vγ−12=300×(11/9)1.5−1=300(3)=900K=627oCOption A.

A fixed amount of dry air at temperature of $27^{o} C$ is compressed to $\frac{1}{9} t h$ of original volume. Its final temperature is $(γ = 1.5)$
$627^{o} C$
$600^{o} C$
$527^{o} C$
$158^{o} C$

In an adiabatic process,
$T V^{γ - 1} =$ constant
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$T_{2} = T_{1} \times \frac{V_{1}^{γ - 1}}{V_{2}^{γ - 1}} = 300 \times {(\frac{1}{1 / 9})}^{1.5 - 1} = 300 (3) = 900 K = 627^{o} C$
Option A.