Question

Figure shows a cylindrical container containing oxygen gas and closed by a piston of mass $50\text{kg}$. Piston can slide smoothly in the cylinder. Its cross sectional area is $100{\text{cm}}^2$ and atmospheric pressure is ${10}^5\text{Pa}$. Some heat is supplied to the cylinder so that the piston is slowly displaced up by $20\text{cm}$.
Find the amount of heat supplied to the gas.

• A. $1050\text{J}$
• B. $750\text{J}$
• C. $105\text{J}$
• D. $1505\text{J}$

Answer 1

The correct option is A $1050\text{J}$

Let us assume the given gas is ideal in nature.
From the figure we can say that, piston is open to atmosphere, thus the gas is under a constant pressure given as
$P_{gas}=P_{atm}+\frac{Mg}{A}$
from the data given in the question,
$P_{gas}={10}^5+\frac{50\times 10}{100\times {10}^{-4}}={10}^5+5\times {10}^4=15\times {10}^4\text{Pa}........\left(1\right)$
When heat is supplied ,it is given that the piston moves out by $\Delta x=20\text{cm}$, thus increment is its volume
$\Delta V=A\times \Delta x$
$=100\times {10}^{-4}\times 20\times {10}^{-2}=2\times {10}^{-3}{\text{m}}^3$
From $\left(1\right)$ we can say that the pressure is constant, so the process is isobaric.
work done by gas during isobaric process is given by
$W=P_{gas}\Delta V$
$=15\times {10}^4\times 2\times {10}^{-3}=300\text{J}$
Since, $W=nR\Delta T=P_{gas}\Delta V$
we can say that, $nR\Delta T=300\text{J}.........\left(2\right)$
Thus heat supplied in raising the temperature by $\Delta T$ is given by
$\Delta Q=n{C}_P\Delta T$
For a diatomic ideal gas, $C_P=\frac{7R}{2}$
$\therefore \Delta Q=n\left(\frac{7}{2}R\right)\Delta T$
From $\left(2\right)$ we get,
$\Delta Q=n\left(\frac{7R}{2}\right)\times \frac{300}{nR}=\frac{7}{2}\times 300=1050\text{J}$
Thus, option (a) is the correct answer.