Question

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 1

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

Explanation :

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

$PV_1=nRT$

where,

P = pressure of gas = 1 atm

$V_1$ = volume of gas = ?

T = temperature of gas = $270^oC=273+270=543K$

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:

$(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K$

$V_1=9.12L$

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

$PV_2=nRT$

where,

P = pressure of gas = 1 atm

$V_2$ = volume of gas = ?

T = temperature of gas = $24^oC=273+24=297K$

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:

$(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K$

$V_2=4.99L$

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 9.12 L

$V_2$ = final volume = 4.99 L

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

$w=-p(V_2-V_1)$

$w=-(1atm)\times (4.99-9.12)L$

$w=4.13L.artm=4.13\times 101.3J=418.4J$

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

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