A van der Waal's gas obeys the equation o | vedclass

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

Question

(c)

Process is quasistatic adiabatic.

$	ext { So, } d Q =0$

$\Rightarrow d W =-d U$

$\Rightarrow p \Delta V =-n C_V d T \quad \dots(i)$

Vedclass · Accepted Answer

$T^{C / n R} \cdot(V-n b)=$ constant

Vedclass · Answer

(c)

Process is quasistatic adiabatic.

$	ext { So, } d Q =0$

$\Rightarrow d W =-d U$

$\Rightarrow p \Delta V =-n C_V d T \quad \dots(i)$

$U =C T-\frac{n^2 a }{V} \quad \dots(ii)$

and $\quad p=\frac{n R T}{V-n b}-\frac{n^2 a}{V^2} \quad \dots(iii)$

From Eq. $(ii)$, $d U=C d T+\frac{n^2 a }{V^2} d V$

So, from Eq. $(i)$, we have

$d W=-d U$

$\Rightarrow \quad p d V=-\left(C d T+\frac{n^2}{V^2} d Vight)$

Substituting for $p$ from Eq. $(iii)$ in above equation, we have

$\Rightarrow\left(\frac{n R T}{V-n b}-\frac{n^2 a}{V^2}ight) d V$

$=-C d T-\frac{n^2 \alpha }{V^2} \cdot d V$

$\Rightarrow\left(\frac{n R T}{V-n b}ight) d V=-C d T$

$\Rightarrow \quad \frac{d V}{V-n b}=\frac{-C}{n R} \cdot \frac{d T}{T}$

$\Rightarrow \quad \int \frac{d V}{V-n b}=-\frac{C}{n R} \int \frac{d T}{T}$

$\Rightarrow \log (V-n b)+k=-\log T^{C / n R}$

$\Rightarrow \log \left(T^{C / n R}  \cdot(V-n b)ight)=$ constant

$\Rightarrow T^{C / n R}  \cdot(V-n b)=$ constant

A van der Waal's gas obeys the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$. Its internal energy is given by $U=C T-\frac{n^2 a}{V}$. The equation of a quasistatic adiabat for this gas is given by

Similar Questions

$Assertion :$ Adiabatic expansion is always accompanied by fall in temperature.
$Reason :$ In adiabatic process, volume is inversely proportional to temperature.

For adiabatic processes $\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)$

If during an adiabatic process the pressure of mixture of gases is found to be proportional to square of its absolute temperature. The ratio of $C_p / C_v$ for mixture of gases is .........

A hypothetical gas expands adiabatically such that its volume changes from $8$ litres to $27$ litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$. Then the ratio of $\frac{C_P}{C_V}$ will be

Neon gas of a given mass expands isothermally to double volume. What should be the further fractional decrease in pressure, so that the gas when adiabatically compressed from that state, reaches the original state?