SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Question : 195 of 200

Marks: +1, -0

In an isentropic process, the pressure of a monoatomic ideal gas increases by 0.5%. The volume will decrease (in %) by (Take

(0.995)^0.625= 0.997):

Solution:

Given:
Pressure increased

=0.5 \%

, So

{P}_{2}=1.005 {P}_{1}

For monoatomic

\γ=1.667

{P}_{1} {V}_{1}^{1.667}={P}_{2} {V}_{2}^{1.667}

{P}_{1} {V}_{1}^{1.667}=1.005 {P}_{1} {V}_{2}^{1.667}

{{P}_{1}}/{1.005 {P}_{1}}=({{V}_{2}}/{{V}_{1}})^{1.667}

({1}/{1.005})^{{1}/{1.667}}=({{V}_{2}}/{{V}_{1}})

(0.995)^{0.6} \× {V}_{1}={V}_{2}

0.997 {V}_{1}={V}_{2}

{V}_{2}=99.7 \%

{V}_{1}

The volume will decrease 0.3%.

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