This is homework problem that i was stuck on ..please show calculations?

Question

A vertical cylinder of cross-sectional area 0.052m^2 is fitting with a tight-fitting, frictionless piston of mass 5.0kg. If there are 3.0 mol of an ideal gas in the cylinder at 600 K, determine the height, h, at which the position will be in equilibrium under its own weight.

Answer 1

here's a start.

Use the ideal gas law.

PV = nRT

details forthcoming, gotta get some lunch first.

ok, now that I have my lunch.

The easiest way to approach this problem is to use unit analysis and maybe a free body diagram.

First, you know that the mass is 50 kg. This means that the force supplied by the pressure in the cylinder is equal to 50 kg * 9.81 kg-m/s^2 (gravittional constant). For simplification I'm going to use 10 instead.

PA = 50 kN (upwards) (Pressure x Area, kN/m^2 * m^2 = kN)

w = 50 kN (downwards)

w = PA, because if it didn't, the piston would be moving.

50 kN = nRT/h

solve for h

h = nRT / 50 kN

I had to look at the units of nRT, or PA to know to divide by h.

Ans. h = 3* 8.314472*600/50000 = .299 m or approx. 30 cm

You could have also broken V into A*h and solved that way.

PAh = nRT...

PA = 50...

50h=nRT

h=nRT/50

Answer 2

sorry, don't know....