A closed vertical cylinder is divided into two parts by a frictionless piston, each part contains 1 mole of air. At $27^{\circ} C$ the volume of the upper part is 4 times than that of the lower part. Calculate the temperature (T) when volume of the upper part will be three times that of the lower part. (write answer to nearest integer function)

Open in App

Solution

The Volume and Temperature are as shown in the above figure.

Initially,
$P_{l o w e r} = P_{u p p e r} + P_{p i s t o n}$
$5 \times \frac{R \times 300}{V} = 5 \times \frac{R \times 300}{4 V} + P_{p i s t o n}$
$P_{p i s t o n} = (\frac{5 R \times 300}{V}) \times \frac{3}{4}$ ............(1)

Finally,
$P_{l o w e r} = P_{u p p e r} + P_{p i s t o n}$

$\frac{4 R T}{V} = \frac{4 R T}{3 V} + P_{p i s t i o n}$

$\frac{8 R T}{3 V} = P_{p i s t o n}$ ...(2)

From equation (1) & and (2)

$\frac{15}{4} \times \frac{R \times 300}{V} = \frac{8 R T}{3 V} \Rightarrow T = 421.9 K$

Answer is $422$ .

Suggest Corrections

Similar questions

Q. A cylindrical container is divided in two equal parts by a diathermic piston. Different ideal gases are filled in the two parts. Find the ratio of the mass of the molecules of the gas in the lower part to that of the upper part, if the root mean square velocity of molecules in the lower part is equal to the mean velocity of molecules in the upper part.

Q. A cone is cut into two parts by a horizontal plane passing through the midpoint of its axis. The ratio of the volume of the upper part to the volume of lower part is ............

Figure shows a vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of the vessel. The length of the cylinder is 90 cm and the piston divides the cylinder in the ratio of 5:4. Each of the two parts of the vessel contains 0.1 mole of an ideal gas. The temperature of the gas is 300 K in each part. Calculate the mass of the piston

A vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of a vessel. The length of vessel is 90 cm and the piston divides the cylinder in the ratio 2 : 1. Each of the two parts contain 0.1 mole of an ideal gas. The temperature of the gas is $27^{o}$ C in each part. The mass of the piston is

Q. A vessel of uniform cross-section of length 500

c m

as shown in figure is divided in two parts by a weightless & frictionless piston. One part contains 5 mole of

N_{2} (g)

and other part contains 2 moles of

H_{2} (g)

& 1 mole of

H e (g)

gaseous mixture at the same temperature and pressure.
What is the length of

N_{2}

compartment?
(Assume volume of piston to be engligible)

Join BYJU'S Learning Program

The Volume and Temperature are as shown in the above figure.Initially,Plower=Pupper+Ppiston5×R×300V=5×R×3004V+PpistonPpiston=(5R×300V)×34............(1)Finally,Plower=Pupper+Ppiston4RTV=4RT3V+Ppistion8RT3V=Ppiston ...(2)From equation (1) & and (2)154×R×300V=8RT3V⇒T=421.9KAnswer is 422.