In an adiabatic process with the ratio of two specific heat, = 3 2 , pressure is increased by 2 3 % , then decrease in the volume will be

Step-by-step solution

For an adiabatic process, the relation between pressure P and volume V is given by PV^ = constant . Taking the natural logarithm on both sides, we get P + V = constant . Differentiating both sides, we obtain dP P + dV V = 0 , which implies dV V = - 1 dP P . Given = 3 2 and the percentage increase in pressure dP P = 2 3 % . Substituting these values, the percentage change in volume is dV V = - 1 3/2 2 3 % = - 2 3 2 3 % = - 4 9 % . The negative sign indicates a decrease in volume. Therefore, the decrease in volume is