Question

The pressure (1 * 10 5 N/m2) of air filled in a vessel is decreased so much adiabatically as to increase its volume three times. Calculate the pressure of air. Gama for air = 1.4 . log10 3 = 0.4771, log 10 2.148 = 0.33206.

Answer 1

Answer:

the pressure of air after its volume is increased three times is

$P_2 = 2.15 \times 10^4 Pa$

Explanation:

As we know by adiabatic process equation

$PV^{\gamma} = C$

so we have

$P_1V_1^{\gamma} = P_2V_2^{\gamma}$

so we have

$P_o(V)^{1.4} = P_2(3V)^{1.4}$

so we have

$P_2 = P_o(\frac{1}{3})^{1.4}$

$P_2 = (1 \times 10^5)(\frac{1}{3})^{1.4}$

$P_2 = 2.15 \times 10^4 Pa$

#Learn

Topic : Adiabatic Process

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