A gas at N.T.P. is suddenly compressed to lef | vedclass

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

Question

(C) For an adiabatic compression,the relation is $PV^{\gamma} = 	ext{constant}$.Given: $V_{	ext{new}} = \frac{1}{4} V$ and $\gamma = \frac{3}{2}$.Us

Vedclass · Accepted Answer

$8\ P$

Vedclass · Answer

(C) For an adiabatic compression,the relation is $PV^{\gamma} = 	ext{constant}$.Given: $V_{	ext{new}} = \frac{1}{4} V$ and $\gamma = \frac{3}{2}$.Using the adiabatic process equation:$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$P \cdot V^{\gamma} = P_{	ext{new}} \cdot \left(\frac{V}{4}ight)^{\gamma}$$\frac{P_{	ext{new}}}{P} = \left(\frac{V}{V/4}ight)^{\gamma} = (4)^{\gamma}$$\frac{P_{	ext{new}}}{P} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$Therefore,$P_{	ext{new}} = 8P$.

$A$ gas at $N.T.P.$ is suddenly compressed to $\left(\frac{1}{4}\right)$ of its original volume. What will be the final pressure? (Given: $\gamma = \text{ratio of specific heats} = \frac{3}{2}$,$P = \text{original pressure}$)

Similar Questions

$A$ motor tube is filled with air at $27^{\circ}C$ and a pressure of $8 \text{ atm}$. If the tube suddenly bursts,what will be the final temperature of the air? $(\gamma = 1.5)$

$A$ gas is being compressed adiabatically. The specific heat of the gas during compression is

For an adiabatic expansion of a perfect gas,the value of $\frac{\Delta P}{P}$ is equal to

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 $(\gamma = \text{ratio of specific heats})$.

In the adiabatic compression, the decrease in volume is associated with

Vedclass Test Series

Exam Paper Generator

Online Exam Module