
A closed hollow insulated cylinder is filled with gas at 0 \[^{0}C\] and also contains an insulated piston of negligible weight and negligible thickness at the middle point. The gas at one side of the piston is heated to 100 \[^{0}C\] . If the piston moves 5cm, the length of the hollow cylinder is
A) 13.65 cm
B) 27.3 cm
C) 38.6 cm
D) 64.6 cm
Answer
565.2k+ views
Hint: The cylinder is closed, so volume does not change during the process. Since the gas is given, we can use ideal gas law. The piston moves so there may be possibility of work done on the system or by the system.
Complete step by step answer:
Ideal gas law states that, for a gas with the number of moles be n, at temperature T and having pressure P and volume V, Using ideal gas law, \[PV=nRT\].
The piston is at the middle point of the container. One side of the gas is at 0\[^{0}C\] and the other is at 100 \[^{0}C\] .
\[\dfrac{PV}{T}=nR\], constant for gas in both sections of the cylinder.
For section with constant temperature, \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
\[\Rightarrow {{P}_{1}}A\times \dfrac{L}{2}={{P}_{2}}A\times (\dfrac{L}{2}-5)\]--------(1)
Now, for the section whose temperature has been increased, using (1)
\[\begin{align}
& \dfrac{{{P}_{1}}A\times \dfrac{L}{2}}{273}=\dfrac{{{P}_{1}}A\times (\dfrac{L}{2}-5)}{373} \\
& L=64.6cm \\
\end{align}\]
So, the correct option is (D).
Additional information- Temperature is a measure of the thermal energy of a body at thermal equilibrium.
Note- We have used ideal gas law here because it was mentioned in the question that the gas is ideal. Otherwise we would have to use real gas laws. Also, while finding volume we have taken the ratio of angles as the radii for both are the same.
Complete step by step answer:
Ideal gas law states that, for a gas with the number of moles be n, at temperature T and having pressure P and volume V, Using ideal gas law, \[PV=nRT\].
The piston is at the middle point of the container. One side of the gas is at 0\[^{0}C\] and the other is at 100 \[^{0}C\] .
\[\dfrac{PV}{T}=nR\], constant for gas in both sections of the cylinder.
For section with constant temperature, \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
\[\Rightarrow {{P}_{1}}A\times \dfrac{L}{2}={{P}_{2}}A\times (\dfrac{L}{2}-5)\]--------(1)
Now, for the section whose temperature has been increased, using (1)
\[\begin{align}
& \dfrac{{{P}_{1}}A\times \dfrac{L}{2}}{273}=\dfrac{{{P}_{1}}A\times (\dfrac{L}{2}-5)}{373} \\
& L=64.6cm \\
\end{align}\]
So, the correct option is (D).
Additional information- Temperature is a measure of the thermal energy of a body at thermal equilibrium.
Note- We have used ideal gas law here because it was mentioned in the question that the gas is ideal. Otherwise we would have to use real gas laws. Also, while finding volume we have taken the ratio of angles as the radii for both are the same.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

