In an H_2 gas process, PV^2 = text{constant}. | vedclass

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

Question

(C) For a polytropic process $PV^x = 	ext{constant}$, the work done is given by $W = \frac{nR \Delta T}{1-x}$.Here, $x = 2$, so $W = \frac{nR \Delta

Vedclass · Accepted Answer

$-0.4$

Vedclass · Answer

(C) For a polytropic process $PV^x = 	ext{constant}$, the work done is given by $W = \frac{nR \Delta T}{1-x}$.Here, $x = 2$, so $W = \frac{nR \Delta T}{1-2} = -nR \Delta T$.For a diatomic gas like $H_2$, the change in internal energy is $\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R ight) \Delta T = 2.5 nR \Delta T$.The ratio of work done to the change in internal energy is $\frac{W}{\Delta U} = \frac{-nR \Delta T}{2.5 nR \Delta T} = -\frac{1}{2.5} = -0.4$.

In an $H_2$ gas process, $PV^2 = \text{constant}$. The ratio of work done by the gas to the change in its internal energy is:

Similar Questions

The variation of pressure $P$ with volume $V$ for an ideal diatomic gas is parabolic as shown in the figure. The molar specific heat of the gas during this process is

An ideal gas expands in such a way that $PV^2 = \text{constant}$ throughout the process.

$P_i, V_i$ and $P_f, V_f$ are initial and final pressures and volumes of a gas in a thermodynamic process respectively. If $PV^n = \text{constant}$, then the amount of work done is

An ideal gas expands in such a manner that its pressure and volume are related by the equation $PV^2 = \text{constant}$. During this process, the gas is

$1 \, \text{mole}$ of gas expands with temperature according to the relation $V = KT^{2/3}$. When the temperature changes by $30 \, ^\circ\text{C}$, the work done will be: (in $, R$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module