Question

$2$ moles of a monatomic ideal gas is enclosed in an adiabatic fixed vertical cylinder fitted with a smooth, light and adiabatic piston. The piston is connected to a vertical spring of spring constant $200\text{N/m}$ as shown in figure. The area of cross-section of the cylinder is $20{\text{cm}}^2$. Initially, the spring is at its natural length and temperature of the gas is $300\text{K}$. The atmospheric pressure is $100\text{kPa}$. The gas is heated slowly for some time by means of an electric heater so as to move the piston up (slowly) by $10\text{cm}$. Then:

• A. Work done by the gas is $21\text{J}$.
• B. Work done on the gas is $21\text{J}$.
• C. Heat supplied by the heater is $772\text{J}$.
• D. Heat supplied by the heater is $793\text{J}$.

Answer 1

Answer: (A), (D)

Heat supplied by the heater is $793\text{J}$.

As the gas is heated and the spring is compressed, the gas expands.
So, work has been done by the gas against atmospheric pressure and the spring force.

Work done by gas $W=\frac{1}{2}k{x}^2+P_0\Delta V$
$=\frac{1}{2}\times 200\times {0.1}^2+{10}^5+20\times {10}^{-4}\times 0.1$
$=21\text{J}$

Also, number of moles of the gas $\left(n\right)=$ constant

$\Rightarrow \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\Rightarrow \frac{{10}^5\times V}{300}=\frac{(P_0+\frac{kx}{A})(V+Ax)}{T_2}----\left(1\right)$

[since $P_{2}A=P_{0}A+kx$ at equilibrium]

From ideal gas equation (initially),
$PV=nRT\Rightarrow {10}^5\times V=2\times 8.3\times 300$
$\Rightarrow V\approx 0.05{\text{m}}^3--\left(2\right)$

From (1) and (2),
$\Rightarrow \frac{{10}^5\times 0.05}{300}=\frac{({10}^5+\frac{200\times 0.1}{20\times {10}^{-4}})(0.05+20\times {10}^{-4}\times 0.1)}{T_2}$
$\Rightarrow \frac{{10}^5\times 0.05}{300}=\frac{5522}{T_2}$
$\Rightarrow T_2\approx 331\text{K}$

Now, change in internal energy,
$\Delta U=\frac{nfR\Delta T}{2}$

$\Rightarrow \Delta U=\frac{2\times 3\times 8.3\times (331-300)}{2}$ [monatomic gas]
$\Rightarrow \Delta U\approx 772\text{J}$

From first law of thermodynamics,
$\Delta Q=\Delta U+W$
$\Rightarrow \Delta Q=772+21=793\text{J}$
(heat supplied by heater)

Hence, options A and D are correct.