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A mixture of gases at $STP$ for which $\gamma=1.5$ is suddenly compressed to $\frac{1}{9}$ th of its original volume. The final temperature of mixture is .......... $^{\circ} C$
A perfect gas goes from state $A$ to another state $B$ by absorbing $8 \times {10^5}J$ of heat and doing $6.5 \times {10^5}J$ of external work. It is now transferred between the same two states in another process in which it absorbs ${10^5}J$ of heat. Then in the second process
One mole of an ideal gas at temperature $T_1$ expends according to the law $\frac{P}{{{V^2}}} =a$ (constant). The work done by the gas till temperature of gas becomes $T_2 $ is
A gas takes part in two processes in which it is heated from the same initial state $1$ to the same final temperature. The processes are shown on the $P-V$ diagram by the straight line $1-2$ and $1-3$. $2$ and $3$ are the points on the same isothermal curve. $Q_1$ and $Q_2$ are the heat transfer along the two processes. Then
The pressure and density of a diatomic gas $(\gamma = 7/5)$ change adiabatically from $(P, d)$ to $(P', d')$. If $\frac{{d'}}{d} = 32$, then $\frac{{P'}}{P}$ should be
Two samples $A$ and $B$ of a gas initially at the same pressure and temperature are compressed from volume $ V$ to $ V/2$ ($A$ isothermally and adiabatically). The final pressure of $ A$ is
A gas mixture consists of $8$ moles of argon and $6$ moles of oxygen at temperature $T$. Neglecting all vibrational modes, the total internal energy of the system is
A monoatomic gas of $n-$moles is heated from temperature $T_1$ to $T_2$ under two different conditions $(i)$ at constant volume and $(ii)$ at constant pressure. The change in internal energy of the gas is