A monoatomic gas is compressed adiabatically | vedclass

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(D) For an adiabatic process,the relation between pressure and volume is given by $PV^\gamma = 	ext{constant}$.Thus,$P_1 V_1^\gamma = P_2 V_2^\gamma$

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$32$ times the initial value

Vedclass · Answer

(D) For an adiabatic process,the relation between pressure and volume is given by $PV^\gamma = 	ext{constant}$.Thus,$P_1 V_1^\gamma = P_2 V_2^\gamma$.Rearranging for the final pressure $P_2$,we get $P_2 = P_1 \left( \frac{V_1}{V_2} ight)^\gamma$.Given that $V_2 = \frac{V_1}{8}$,so $\frac{V_1}{V_2} = 8$.Substituting the values,$P_2 = P_1 (8)^{5/3}$.Since $8 = 2^3$,we have $P_2 = P_1 (2^3)^{5/3} = P_1 (2^5) = 32 P_1$.Therefore,the pressure becomes $32$ times the initial value.

$A$ monoatomic gas is compressed adiabatically to $1/8$ of its initial volume. What happens to the gas pressure? $(\gamma = 5/3)$

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