One mole of an ideal diatomic gas undergoes a | vedclass

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

Question

(A) For an ideal gas,the change in internal energy is given by $\Delta U = n C_V \Delta T$.Since $C_V = \frac{R}{\gamma - 1}$ and $PV = nRT$,we have $

Vedclass · Accepted Answer

$-\,20\,kJ$

Vedclass · Answer

(A) For an ideal gas,the change in internal energy is given by $\Delta U = n C_V \Delta T$.Since $C_V = \frac{R}{\gamma - 1}$ and $PV = nRT$,we have $\Delta U = \frac{n R \Delta T}{\gamma - 1} = \frac{P_B V_B - P_A V_A}{\gamma - 1}$.For a diatomic gas,the adiabatic index $\gamma = 1.4 = \frac{7}{5}$,so $\gamma - 1 = 0.4 = \frac{2}{5}$.From the graph,at point $A$: $P_A = 5 	imes 10^3 \, Pa$,$V_A = 4 \, m^3$.At point $B$: $P_B = 2 	imes 10^3 \, Pa$,$V_B = 6 \, m^3$.Calculating the product $PV$ at each point:$P_A V_A = (5 	imes 10^3) 	imes 4 = 20 	imes 10^3 \, J$.$P_B V_B = (2 	imes 10^3) 	imes 6 = 12 	imes 10^3 \, J$.Substituting these values into the formula:$\Delta U = \frac{12 	imes 10^3 - 20 	imes 10^3}{2/5} = \frac{-8 	imes 10^3}{0.4} = -20 	imes 10^3 \, J = -20 \, kJ$.

One mole of an ideal diatomic gas undergoes a transition from $A$ to $B$ along a path $AB$ as shown in the figure. The change in internal energy of the gas during the transition is:

Similar Questions

$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to (in $\pi$)

Which kingdom is characterized by the presence of a rigid cell wall and the formation of an embryo?

If $y=y(x)$ is the solution of the differential equation $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$ with $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

$A$ stone thrown upwards has its equation of motion $s = 490t - 4.9t^2$. Then the maximum height reached by it is

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module