How much should the pressure be increased in | vedclass

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

Question

(C) According to Boyle's Law,for a given mass of gas at a constant temperature,$P_1 V_1 = P_2 V_2$.Let the initial pressure be $P_1 = P$ and the initi

Vedclass · Accepted Answer

$5.26$

Vedclass · Answer

(C) According to Boyle's Law,for a given mass of gas at a constant temperature,$P_1 V_1 = P_2 V_2$.Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$.The volume is reduced by $5 \%$,so the final volume $V_2 = V - 0.05V = 0.95V$.Substituting these into the equation: $P 	imes V = P_2 	imes 0.95V$.Solving for $P_2$: $P_2 = P / 0.95 = (100/95)P = (20/19)P \approx 1.0526P$.The increase in pressure is $\Delta P = P_2 - P = 1.0526P - P = 0.0526P$.The percentage increase in pressure is $(\Delta P / P) 	imes 100 = 0.0526 	imes 100 = 5.26 \%$.

How much should the pressure be increased in order to reduce the volume of a given mass of gas by $5 \%$ at a constant temperature (in $\%$)?

Similar Questions

The gas equation $\frac{PV}{T} = \text{constant}$ is true for a constant mass of an ideal gas undergoing:

What is the equation of state for $8 \, g$ of $O_2$?

One mole of a gas at a pressure $2 \ Pa$ and temperature $27^{\circ} C$ is heated until both pressure and volume are doubled. What is the final temperature of the gas (in $K$)?

The temperature of an ideal gas at atmospheric pressure is $300\,K$ and volume is $1\,m^3$. If the temperature and volume both become double,then the pressure will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module