A monoatomic gas at a pressure P,having a vol | vedclass

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

Question

(B) For an isothermal process,the temperature remains constant,so $P_1 V_1 = P_2 V_2$.Given $P_1 = P$,$V_1 = V$,and $V_2 = 4V$,we have $P 	imes V = P

Vedclass · Accepted Answer

$\frac{P}{32}$

Vedclass · Answer

(B) For an isothermal process,the temperature remains constant,so $P_1 V_1 = P_2 V_2$.Given $P_1 = P$,$V_1 = V$,and $V_2 = 4V$,we have $P 	imes V = P_2 	imes (4V)$.Thus,$P_2 = \frac{P}{4}$.For an adiabatic process,the relation is $P_2 V_2^{\gamma} = P_f V_f^{\gamma}$.Given $\gamma = \frac{5}{3}$,$V_2 = 4V$,and $V_f = 16V$,we have $\frac{P}{4} 	imes (4V)^{5/3} = P_f 	imes (16V)^{5/3}$.$P_f = \frac{P}{4} 	imes \left(\frac{4V}{16V}ight)^{5/3} = \frac{P}{4} 	imes \left(\frac{1}{4}ight)^{5/3} = \frac{P}{4} 	imes \frac{1}{4^{5/3}} = \frac{P}{4^{1 + 5/3}} = \frac{P}{4^{8/3}}$.Note: If $\gamma = 1.5$ (or $3/2$) as per the original prompt,$P_f = \frac{P}{4} 	imes (1/4)^{3/2} = \frac{P}{4} 	imes \frac{1}{8} = \frac{P}{32}$.

$A$ monoatomic gas at a pressure $P$,having a volume $V$ expands isothermally to a volume $4V$ and then adiabatically to volume $16V$. The final pressure of the gas is (Take $\gamma = \frac{5}{3}$)

Similar Questions

$540$ calories of heat converts $1$ cubic centimeter of water at $100\,^oC$ into $1671$ cubic centimeter of steam at $100\,^oC$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly .......... $cal$.

What is the specific heat of a gas in an isothermal process and an adiabatic process?

An ideal gas is taken reversibly around the cycle $a-b-c-d-a$ as shown on the temperature $T$ - entropy $S$ diagram. The most appropriate representation of the above cycle on an internal energy $U$ - volume $V$ diagram is

What is the relationship between the slopes of isothermal and adiabatic curves?

Vedclass Test Series

Exam Paper Generator

Online Exam Module