An ideal gas is enclosed in a container of vo | vedclass

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

Question

(A) Since the temperature remains constant, we use Boyle's Law, $PV = 	ext{constant}$.Initially, the gas occupies volume $V$ at pressure $P$. When th

Vedclass · Accepted Answer

$P\left(\frac{V}{V+v}\right)^n$

Vedclass · Answer

(A) Since the temperature remains constant, we use Boyle's Law, $PV = 	ext{constant}$.Initially, the gas occupies volume $V$ at pressure $P$. When the pump performs one stroke, the gas expands to fill the total volume $(V + v)$.Let $P_1$ be the pressure after the first stroke:$P \cdot V = P_1 \cdot (V + v)$$P_1 = P \left( \frac{V}{V + v} ight)$After the second stroke, the gas at pressure $P_1$ in volume $V$ expands to volume $(V + v)$:$P_1 \cdot V = P_2 \cdot (V + v)$$P_2 = P_1 \left( \frac{V}{V + v} ight) = P \left( \frac{V}{V + v} ight)^2$Following this pattern, after $n$ strokes, the pressure $P_n$ will be:$P_n = P \left( \frac{V}{V + v} ight)^n$

An ideal gas is enclosed in a container of volume $V$ at a pressure $P$. It is being pumped out of the container by using a pump with stroke volume $v$. What is the final pressure in the container after $n$ strokes of the pump? (Assume temperature remains constant.)

Similar Questions

State Boyle's Law and Charles's Law.

An open-mouthed container is filled with air at $60^{\circ}C$. The container is heated to a temperature $T$ such that $1/4$ of the air escapes. The value of $T$ is ........ $^{\circ}C$.

The pressure-temperature graph for an ideal gas is shown in the figure. At point $A$,the density of the gas is $\rho_0$. What will be the density at point $B$?

At constant temperature,increasing the pressure of a gas by $5\%$ will decrease its volume by ..... $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module