In an adiabatic process,if the pressure is in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Taking the natural logarithm

Vedclass · Accepted Answer

$\frac{4}{9}\% $

Vedclass · Answer

(A) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Taking the natural logarithm on both sides: $\ln P + \gamma \ln V = 	ext{constant}$.Differentiating both sides: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.This implies $\frac{dP}{P} = -\gamma \frac{dV}{V}$.Given $\frac{dP}{P} = \frac{2}{3}\% $ and $\gamma = \frac{3}{2}$.Substituting these values: $\frac{2}{3}\% = -\frac{3}{2} \frac{dV}{V}$.Solving for the percentage change in volume: $\frac{dV}{V} = -\frac{2}{3} 	imes \frac{2}{3}\% = -\frac{4}{9}\%$.The negative sign indicates a decrease in volume. Thus,the volume decreases by $\frac{4}{9}\% $.

In an adiabatic process,if the pressure is increased by $\frac{2}{3}\% $ and the ratio of specific heats $\frac{C_p}{C_v} = \gamma = \frac{3}{2}$,then the volume decreases by about:

Similar Questions

If during an adiabatic process the pressure of a mixture of gases is found to be proportional to the square of its absolute temperature,the ratio of $C_p / C_v$ for the mixture of gases is .........

One mole of an ideal gas $(\gamma = 1.4)$ is adiabatically compressed so that its temperature rises from $27\,^{\circ}C$ to $35\,^{\circ}C$. The change in the internal energy of the gas is .... $J$ (given $R = 8.3 \,J/mol/K$)

In an adiabatic expansion of a gas,

If $5.6$ litres of a monoatomic gas at $STP$ is adiabatically compressed to $0.7$ litres,then the work done on the gas is nearly ($R$ = Universal gas constant). (in $R$)

What is the name of the ideal-gas process in which no heat is transferred?

Vedclass Test Series

Exam Paper Generator

Online Exam Module