An ideal gas is taken through a quasi-static process described by $P = \alpha\, V^2$, with $\alpha = 5\,atm/m^6$. The gas is expanded to twice its original volume of $1\,m^3$. How much work is done by the expanding gas in this process
A$10.8\,J$
B$1.18\,M\,J$
C$5\,J$
D$4.3\,M\,J$
Medium
Download our app for free and get started
B$1.18\,M\,J$
b $\mathrm{W}=\int_{\mathrm{V}_{1}}^{\mathrm{v}_{2}} \mathrm{P} \mathrm{d} \mathrm{v}$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
An ideal gas undergoes an adiabatic process obeying the relation $PV^{4/3} =$ constant. If its initial temperature is $300\,\, K$ and then its pressure is increased upto four times its initial value, then the final temperature is (in Kelvin):
Given that $1\,g$ of water in liquid phase has volume $1\,cm^3$ and in vapour phase $1671\, cm^3$ at atmospheric pressure and the latent heat of vaporization of water is $2256\,J/g;$ the change in the internal energy in joules for $1\,g$ of water at $373\,K$ when it changes from liquid phase to vapour phase at the same temperature is ....... $J$
Under isothermal condition, the pressure of a gas is given by $P = aV ^{-3}$, where $a$ is a constant and $V$ is the volume of the gas. The bulk modulus at constant temperature is equal to $..........\,P$
The efficiency of the cycle shown below in the figure (consisting of one isobar, one adiabat and one isotherm) is $50 \%$. The ratio $x$, between the highest and lowest temperatures attained in this cycle obeys (the working substance is an ideal gas)
A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $\left(\right.$ given, $\left.\gamma=\frac{3}{2}\right)$ :