5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be $T_1$, the work done in the process is,
Solution: For an ideal gas, $pV=nRT$. In an adiabatic process, $TV^{\gamma-1}=\text{constant}$, and the work done by the gas is given by, \begin{alignat}{2} W&=\frac{p_1 V_1 -p_2 V_2}{\gamma -1} \\ &=\frac{nRT_1 -nRT_2}{\gamma -1} \\ &=\frac{nRT_1}{\gamma -1}\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right], \nonumber \end{alignat} where $V_1=5.6\;\mathrm{litre}$ and $V_2=0.7\;\mathrm{litre}$ are initial and final volume of the gas. Now, 5.6 litre at STP gives $n=5.6/22.4=1/4\;\mathrm{mol}$. Helium, being a monoatomic gas, has $\gamma=5/3$. Substitute these values to get $W=-\frac{9}{8}RT_1$. Negative sign indicates that work is done on the gas.
Starting with the same initial conditions, an ideal gas expands from volume $V_1$ to $V_2$ in three different ways. The work done by the gas is $W_1$ if the process is purely isothermal, $W_2$ if purely isobaric and $W_3$ if purely adiabatic, then,
Solution: Let $p_1$, $V_1$, and $T_1$ be initial pressure, volume, and temperature of the gas. For an ideal gas, $p_1V_1=nRT_1$. The work done by the gas in isothermal, isobaric, and adiabatic processes are given by, \begin{align} W_1&=p_1V_1\ln \left(\frac{V_2}{V_1}\right), \nonumber\\ W_2&=p_1(V_2-V_1), \nonumber\\ W_3&=\frac{p_1V_1-p_2V_2}{\gamma-1} \\ &=\frac{p_1V_1}{\gamma-1}\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right], \nonumber \end{align} where $p_2$ and $V_2$ are final pressure and volume. If relationship among $W_1$, $W_2$, and $W_3$ holds in general then it should hold good for any special case, say $V_2=2V_1$ and $\gamma=2$. In this case, $W_1=0.693 p_1 V_1$, $W_2=p_1 V_1$, and $W_3=0.5 p_1 V_1$, giving $W_2 > W_1 > W_3$.
An ideal gas expands isothermally from a volume $V_1$ to $V_2$ and then compressed to original volume $V_1$ adiabatically. Initial pressure is $p_1$ and final pressure is $p_3$. The total work done is $W$. Then,
Solution: On a $p\text{-}V$ diagram, magnitude of slope of an adiabatic process is greater than that of an isothermal process (adiabatic process is more steeper). This fact is applied at the point B to draw the isothermal process AB and the adiabatic process BC (see figure).
Question 1: Which of the following is not true for an isobaric process?
Question 2: Which of the following is not true for an isochoric process?
Question 3: An isothermal and an adiabatic processes are shown on a p-V diagram. These processes intersect at a point A. The magnitude of the slope of adiabatic process at A is more than the magnitude of slope of isothermal process at A.