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

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