In an adiabatic process, the density of a dia | vedclass

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

Question

In adiabatic process

$PV ^{\gamma}= constant$

$P \left(\frac{ m }{\rho}\right)^{\gamma}= constant$

as mass is constant

$P \propto \rho^{\g

Vedclass · Accepted Answer

$128$

Vedclass · Answer

In adiabatic process

$PV ^{\gamma}= constant$

$P \left(\frac{ m }{\rho}\right)^{\gamma}= constant$

as mass is constant

$P \propto \rho^{\gamma}$

$\frac{P_{f}}{P_{i}}=\left(\frac{\rho_{f}}{\rho_{i}}\right)^{\gamma}=(32)^{7 / 5}=2^{7}=128$

List$-I$	List$-II$
$(a)$ Isothermal	$(i)$ Pressure constant
$(b)$ Isochoric	$(ii)$ Temperature constant
$(c)$ Adiabatic	$(iii)$ Volume constant
$(d)$ Isobaric	$(iv)$ Heat content is constant

In an adiabatic process, the density of a diatomic gas becomes $32$ times its initial value. The final pressure of the gas is found to be $n$ times the initial pressure. The value of $n$ is

Similar Questions

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

If $\gamma $ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal $PV$ curves at their point of intersection is

Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, are compressed till their volume becomes half the initial volume.

The work of $146\ kJ$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $7^o C$. The gas is $(R=8.3\ J\ mol^{-1} K^{-1})$

Match List$-I$ with List$-II$

List$-I$ List$-II$

$(a)$ Isothermal $(i)$ Pressure constant

$(b)$ Isochoric $(ii)$ Temperature constant

$(c)$ Adiabatic $(iii)$ Volume constant

$(d)$ Isobaric $(iv)$ Heat content is constant

Choose the correct answer from the options given below