For an ideal gas,the instantaneous change in | vedclass

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

Question

(A) Given the differential equation $\frac{dp}{dv} = -ap$. Integrating both sides: $\int_{p_{0}}^{p} \frac{dp}{p} = -a \int_{0}^{v} dv$. This yields $

Vedclass · Accepted Answer

$\frac{p_{0}}{aeR}$

Vedclass · Answer

(A) Given the differential equation $\frac{dp}{dv} = -ap$. Integrating both sides: $\int_{p_{0}}^{p} \frac{dp}{p} = -a \int_{0}^{v} dv$. This yields $\ln(\frac{p}{p_{0}}) = -av$,so $p = p_{0}e^{-av}$. For one mole of an ideal gas,$PV = RT$,so $T = \frac{PV}{R} = \frac{p_{0}ve^{-av}}{R}$. To find the maximum temperature,differentiate $T$ with respect to $v$ and set it to zero: $\frac{dT}{dv} = \frac{p_{0}}{R} [e^{-av} + v(-a)e^{-av}] = \frac{p_{0}e^{-av}}{R} (1 - av) = 0$. This gives $v = \frac{1}{a}$. Substituting $v = \frac{1}{a}$ into the temperature equation: $T_{max} = \frac{p_{0}(\frac{1}{a})e^{-a(\frac{1}{a})}}{R} = \frac{p_{0}}{aeR}$.

For an ideal gas,the instantaneous change in pressure $p$ with volume $v$ is given by the equation $\frac{dp}{dv} = -ap$. If $p = p_{0}$ at $v = 0$ is the given boundary condition,then the maximum temperature one mole of gas can attain is: (Here $R$ is the gas constant)

Similar Questions

When the gas is heated,in which direction will the piston move?

$A$ $15\,g$ mass of nitrogen gas is enclosed in a vessel at a temperature $27\,^{\circ}C.$ The amount of heat transferred to the gas,so that the rms velocity of the molecules is doubled,is about ...... $kJ$ [Take $R = 8.3\,J/K\,mole$ ]

Molecules in the atmosphere are attracted by the gravitational force of the Earth. Explain why all of them do not fall onto the Earth,just like an apple falling from a tree.

The mass of a hydrogen atom is $3.32 \times 10^{-27} \, kg$. If $10^{23}$ atoms per second strike a wall of area $2 \, cm^2$ at an angle of $45^{\circ}$ with the normal to the wall and reflect elastically with a speed of $10^3 \, m/s$,the pressure exerted on the wall will be: $(N/m^2)$

The average translational kinetic energy of ${N}_{2}$ gas molecules at $... {}^{\circ} {C}$ becomes equal to the ${K.E.}$ of an electron accelerated from rest through a potential difference of $0.1 \ V$. (Given ${k}_{B} = 1.38 \times 10^{-23} \ J/K$). (Fill the nearest integer).

Vedclass Test Series

Exam Paper Generator

Online Exam Module