A heating element of resistance r is fitted i | vedclass

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

Question

(A, C) For a diatomic gas,the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta t

Vedclass · Accepted Answer

the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$

Vedclass · Answer

(A, C) For a diatomic gas,the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta t^2)$.The rate of increase in internal energy is $\frac{dU}{dt} = \frac{5 R}{2} (\alpha + \beta t)$.Since the pressure is constant,the heat supplied is $dQ = n C_p dT$. For a diatomic gas,$C_p = \frac{7}{2} R$.Thus,the power supplied by the heating element is $i^2 r = \frac{dQ}{dt} = n C_p \frac{dT}{dt} = 1 	imes \frac{7}{2} R 	imes (\alpha + \beta t)$.Therefore,the current $i = \sqrt{\frac{7 R}{2 r} (\alpha + \beta t)}$.From the ideal gas law $PV = nRT$,since $P$ is constant,$V = \frac{nRT}{P} = \frac{R}{P} (\alpha t + \frac{1}{2} \beta t^2)$.The position of the piston $x = \frac{V}{A} = \frac{R}{PA} (\alpha t + \frac{1}{2} \beta t^2)$.The velocity $v = \frac{dx}{dt} = \frac{R}{PA} (\alpha + \beta t)$.The acceleration $a = \frac{dv}{dt} = \frac{R \beta}{PA}$,which is constant. Thus,options $(a)$ and $(c)$ are correct.

Similar Questions

The internal energy of an ideal diatomic gas corresponding to volume $V$ and pressure $P$ is $2.5 PV$. The gas expands from $1 \text{ litre}$ to $2 \text{ litre}$ at a constant pressure of $10^5 \text{ N/m}^2$. The heat supplied to the gas is: (in $\text{ J}$)

$A$ quantity of heat $Q$ is supplied to a monoatomic ideal gas which expands at constant pressure. What is the fraction of heat converted into work? Given $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$.

The slopes of isothermal and adiabatic curves are related as

When a system is taken from state $i$ to state $f$ along the path $iaf$,$Q = 50 \ J$ and $W = 20 \ J$. Along the path $ibf$,$Q = 35 \ J$ and $W = -13 \ J$. For the curved path $fi$,$Q = \dots \ J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module