A cylinder with adiabatic walls is closed at | vedclass

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

Question

(B) The right side compartment undergoes an adiabatic compression because the piston is adiabatic and frictionless.For an adiabatic process, the relat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The right side compartment undergoes an adiabatic compression because the piston is adiabatic and frictionless.For an adiabatic process, the relation is $PV^{\gamma} = 	ext{constant}$.Given $\gamma = 1.5 = 3/2$.Let the initial state be $(P, V_1 = 9)$ and the final state be $(P_2 = \frac{27}{8}P, V_2)$.Since the piston is frictionless, the pressure on both sides must be equal at equilibrium. Thus, the final pressure on the right side is also $\frac{27}{8}P$.Using the adiabatic relation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.$P(9)^{3/2} = \frac{27}{8}P(V_2)^{3/2}$.$(9)^{3/2} = \frac{27}{8} V_2^{3/2}$.$27 = \frac{27}{8} V_2^{3/2}$.$V_2^{3/2} = 8$.$V_2 = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4 	ext{ litres}$.

Similar Questions

In a thermodynamic process,there is no exchange of heat between the system and surroundings. Then the thermodynamic process is

Can the temperature of an isolated system of gas be changed?

$A$ monoatomic gas $\left( \gamma = \frac{5}{3} \right)$ is suddenly compressed to $\frac{1}{8}$ of its original volume. What will be the final pressure of the gas in terms of its initial pressure?

The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

The temperature of a hypothetical gas increases to $\sqrt{2}$ times when compressed adiabatically to half the volume. Its equation can be written as

Vedclass Test Series

Exam Paper Generator

Online Exam Module