The ratio of specific heats of a gas is gamma | vedclass

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

Question

(C) The change in internal energy for an ideal gas is given by $\Delta U = n C_V \Delta T$. For one mole of gas $(n=1)$,$\Delta U = C_V \Delta T$. We

Vedclass · Accepted Answer

$\frac{p V}{\gamma-1}$

Vedclass · Answer

(C) The change in internal energy for an ideal gas is given by $\Delta U = n C_V \Delta T$. For one mole of gas $(n=1)$,$\Delta U = C_V \Delta T$. We know that $C_V = \frac{R}{\gamma-1}$. Thus,$\Delta U = \frac{R \Delta T}{\gamma-1}$. From the ideal gas equation at constant pressure $p$,$p V = R T$,so $p \Delta V = R \Delta T$. Here,the change in volume is $\Delta V = 2V - V = V$. Therefore,$R \Delta T = p V$. Substituting this into the expression for $\Delta U$: $\Delta U = \frac{p V}{\gamma-1}$.

The ratio of specific heats of a gas is $\gamma$. The change in internal energy of one mole of the gas,when the volume changes from $V$ to $2 V$ at constant pressure $p$ is

Similar Questions

If the magnitudes of $a, b$ and $a+b$ are respectively $3, 4$ and $5$,then the magnitude of $(a-b)$ is

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

The first minima due to a single slit diffraction is at $\theta = 30^o$ for a light of wavelength $5000\,\mathring{A}$. The width of the slit is:

When a body moves in a circular path,no work is done by the force since,

The product of the lengths of the perpendiculars from any point on the hyperbola $x^2 - y^2 = 8$ to its asymptotes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module