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svp [43]
3 years ago
5

Air compressed in car engine from 25 C and 100 kPa in reversible and adiabatic manner. If the compression ratio = 9.058, determi

ne final temperature of air.
Physics
1 answer:
prisoha [69]3 years ago
3 0

Explanation:

It is given that,

Initial temperature, T_1=25^{\circ}C=298\ K

Pressure, P_1=100\ kPa=10^5\ Pa

Compression ratio, r=\dfrac{V_1}{V_2}=9.058

Let T₂ is the final temperature of air. Using the relation for reversible adiabatic process as :

\dfrac{T_2}{T_1}=(\dfrac{V_1}{V_2})^{\gamma-1}............(1)

Where,

\gamma=\dfrac{C_p}{C_v}

For air, C_p=1.004 and C_v=0.717

\gamma=1.4

So, equation (1) becomes :

T_2=T_1\times (\dfrac{V_1}{V_2})^{\gamma-1}

T_2=298\times (9.058)^{1.4-1}

T_2=719.49\ K

So, the final temperature of air is 719.49 K. Hence, this is the required solution.

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