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](https://tex.z-dn.net/?f=PV_1%3DnRT)
where,
P = pressure of gas = 1 atm
= volume of gas = ?
T = temperature of gas = ![270^oC=273+270=543K](https://tex.z-dn.net/?f=270%5EoC%3D273%2B270%3D543K)
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](https://tex.z-dn.net/?f=%281atm%29%5Ctimes%20V_1%3D0.205mol%5Ctimes%200.0821L.atm%2Fmol.K%5Ctimes%20543K)
![V_1=9.12L](https://tex.z-dn.net/?f=V_1%3D9.12L)
Now we have to calculate the volume at 24°C.
![PV_2=nRT](https://tex.z-dn.net/?f=PV_2%3DnRT)
where,
P = pressure of gas = 1 atm
= volume of gas = ?
T = temperature of gas = ![24^oC=273+24=297K](https://tex.z-dn.net/?f=24%5EoC%3D273%2B24%3D297K)
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](https://tex.z-dn.net/?f=%281atm%29%5Ctimes%20V_2%3D0.205mol%5Ctimes%200.0821L.atm%2Fmol.K%5Ctimes%20297K)
![V_2=4.99L](https://tex.z-dn.net/?f=V_2%3D4.99L)
Now we have to calculate the work done.
Formula used :
![w=-p\Delta V\\\\w=-p(V_2-V_1)](https://tex.z-dn.net/?f=w%3D-p%5CDelta%20V%5C%5C%5C%5Cw%3D-p%28V_2-V_1%29)
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:
![w=-p(V_2-V_1)](https://tex.z-dn.net/?f=w%3D-p%28V_2-V_1%29)
![w=-(1atm)\times (4.99-9.12)L](https://tex.z-dn.net/?f=w%3D-%281atm%29%5Ctimes%20%284.99-9.12%29L)
![w=4.13L.artm=4.13\times 101.3J=418.4J](https://tex.z-dn.net/?f=w%3D4.13L.artm%3D4.13%5Ctimes%20101.3J%3D418.4J)
conversion used : (1 L.atm = 101.3 J)
Therefore, the work done on the gas will be, 418.4 J