A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder, how much work will be done on the gas when its temperature is decreased from 270 ∘C to 24∘C? Hint: Since the piston slides without friction, the pressure will remain constant. Use the ideal gas law to change W=−PΔ to an expression using and T. Be sure to include the appropriate sign of the work in the answer.

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

where,

P = pressure of gas = 1 atm

= volume of gas = ?

T = temperature of gas =

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the volume at 24°C.

where,

P = pressure of gas = 1 atm

= volume of gas = ?

T = temperature of gas =

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the work done.

Formula used :

where,

w = work done

p = pressure of the gas = 1 atm

= initial volume = 9.12 L

= final volume = 4.99 L

Now put all the given values in the above formula, we get:

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

to make a supersaturated solution, make a saturated solution of sugar by adding 360 grams of sugar to 100 ml of water at 80 degrees celsius. when the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water should only dissolve 210 grams of sugar.