Neon gas of a given mass expands isothermally | vedclass

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

Question

(A) For an isothermal process,$P_1 V_1 = P_2 V_2$. Given $V_2 = 2V_1$,we have $P_1 V_1 = P_2 (2V_1)$,which gives $P_2 = P_1 / 2$.Let the pressure be f

Vedclass · Accepted Answer

$1-2^{-2 / 3}$

Vedclass · Answer

(A) For an isothermal process,$P_1 V_1 = P_2 V_2$. Given $V_2 = 2V_1$,we have $P_1 V_1 = P_2 (2V_1)$,which gives $P_2 = P_1 / 2$.Let the pressure be further decreased by a factor such that the new pressure is $P'$. We want to reach the original state $(P_1, V_1)$ from $(P', 2V_1)$ via an adiabatic process.For an adiabatic process,$P' (2V_1)^\gamma = P_1 (V_1)^\gamma$,where $\gamma = 5/3$ for a monoatomic gas like neon.$P' (2)^\gamma = P_1 \implies P' = P_1 (2)^{-\gamma} = P_1 (2)^{-5/3}$.The pressure at the end of the isothermal expansion was $P_2 = P_1 / 2 = P_1 (2)^{-1}$.The fractional decrease in pressure required is $\frac{P_2 - P'}{P_2} = \frac{P_1 (2)^{-1} - P_1 (2)^{-5/3}}{P_1 (2)^{-1}} = 1 - \frac{(2)^{-5/3}}{(2)^{-1}} = 1 - 2^{-5/3 + 1} = 1 - 2^{-2/3}$.

Neon gas of a given mass expands isothermally to double its volume. What should be the further fractional decrease in pressure,so that the gas when adiabatically compressed from that state,reaches the original state?

Similar Questions

The efficiency of a heat engine is $\eta$ and the coefficient of performance of a refrigerator is $\beta$. Then:

An ideal monoatomic gas is carried along the cycle $ABCDA$ as shown in the figure. The total heat absorbed during this process is (in $p_0 V_0$)

Select the correct statement for work, heat, and change in internal energy.

What are thermodynamic processes?

The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is

Vedclass Test Series

Exam Paper Generator

Online Exam Module