The P-V diagram for two gases during an adiab | vedclass

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

Question

(A) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} +

Vedclass · Accepted Answer

$O_2$ and $He$

Vedclass · Answer

(A) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} + P \gamma V^{\gamma-1} = 0$.Thus, the slope $m = \frac{dP}{dV} = -\gamma \frac{P}{V}$.From the figure, the magnitude of the slope of curve $B$ is greater than the magnitude of the slope of curve $A$ $(|m_B| > |m_A|)$.Therefore, $\gamma_B > \gamma_A$.For monoatomic gases (like $He$, $Ar$), $\gamma = 1.67$.For diatomic gases (like $O_2$), $\gamma = 1.4$.Since $\gamma_B > \gamma_A$, curve $B$ represents a monoatomic gas and curve $A$ represents a diatomic gas.Thus, $A$ corresponds to $O_2$ and $B$ corresponds to $He$ (or $Ar$).Comparing with the options, $A$ is $O_2$ and $B$ is $He$ fits the condition.

The $P-V$ diagram for two gases during an adiabatic process is shown in the figure. Curves $A$ and $B$ correspond to:

Similar Questions

The internal energy of the gas increases in:

An ideal gas expands adiabatically, $(\gamma = 1.5)$. To reduce the root-mean-square (r.m.s.) velocity of the molecules $4$ times, the gas has to be expanded by a factor of: (in $\times$)

Five moles of Hydrogen gas initially at $STP$ is compressed adiabatically so that its temperature becomes $673 \, K$. The increase in internal energy of the gas is $(R=8.3 \, J \, mol^{-1} \, K^{-1}, \gamma=1.4$ for diatomic gas$)$ (in $kJ$)

In the adiabatic compression, the decrease in volume is associated with

$\Delta U + \Delta W = 0$ is valid for

Vedclass Test Series

Exam Paper Generator

Online Exam Module