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Question: One mole of gas occupying 3 liters of volume is expanded against constant external pressure of one a...

One mole of gas occupying 3 liters of volume is expanded against constant external pressure of one atm to a volume of 15 liters. The work done by the system is:
(A) −1.215×103 - 1.215 \times {10^3} J
(B) +12.15×103 + 12.15 \times {10^3} J
(C) +121.5×103 + 121.5 \times {10^3} J
(D) +1.215×103 + 1.215 \times {10^3} J

Explanation

Solution

In order to solve these types of questions we should first recognize the type of process and according to which we will be knowing the work done in different processes is different. Since pressure is constant it is an isobaric process. And work done in isobaric process is equal to PΔVP\Delta V

Complete answer: For solving this question we need to know what kind of process so this can be decoded by the help of information given in the question since it is given in question the volume is expanded against constant external pressure, so we noted that the pressure is constant so it is an isobaric process.Now we have to learn about isobaric process:The isobaric process occurs at constant pressure. Since the pressure is constant the force exerted is constant and work done is given as: PΔVP\Delta V
And the whole process is called isobaric expansion.Since the force applied is external force so the work done will be negative.So the equation will be −PΔV - P\Delta V
Up till now it is clear now that the correct option will be A but we need to solve this.So the work done by the system against an external pressure is given byW=−Pext.ΔVW = - {P_{ext.}}\Delta V
Now we will find what are the resource given in question can be utilized:
Pext.{P_{ext.}} = 1 atm
Vf{V_f} = 15L
Vi{V_i} = 3L
So according to the formula putting all the values in it:
W=−1(15−3)W = - 1(15 - 3)
W=−12LatmW = - 12L atm
Since the options are given in Joules so we need to convert it in Joules.
Since we know that 1Latm=101.325J1L atm = 101.325J
Therefore,−12×101.325=−1.215×103 - 12 \times 101.325 = - 1.215 \times {10^3}
Hence the work done by the system is −1.215×103 - 1.215 \times {10^3} J.

So,(A) option is correct.

Note: However there are more such three processes rather than isobaric which are isochoric in which volume is constant, isothermal in which temperature is constant and adiabatic in which no heat is transferred (exchanged) by the system.