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Question: One mole of gas occupying \(3d{{m}^{3}}\) expands against a constant external pressure of 1 atm to a...

One mole of gas occupying 3dm33d{{m}^{3}} expands against a constant external pressure of 1 atm to a volume of 13 lit. The work done is
(1) −10 atm dm3-10\,atm\,d{{m}^{3}}
(2) −20 atm dm3-20\,atm\,d{{m}^{3}}
(3) −39 atm dm3-39\,atm\,d{{m}^{3}}
(4) −48 atm dm3-48\,atm\,d{{m}^{3}}

Explanation

Solution

To solve this question we have to use the formula of work done for irreversible processes. The process is irreversible because the work is done against the constant external pressure. The formula for work done is given as W=−PextΔVW=-{{P}_{ext}}\Delta V, where Pext{{P}_{ext}}is the external pressure and ΔV\Delta V is the change in volume.

Complete step by step solution:
From your chemistry lessons you have learned about the isobaric process where pressure remains constant and the formula of work done when pressure is kept constant. An isobaric process is defined as the thermodynamic process which takes place at constant pressure. The term isobaric is given through a Greek words where "iso" and "baro" means equal pressure. When the volume is expanded or contracted the constant pressure is obtained. Isobaric process neutralizes any pressure change which arises due to the transfer of heat. So, when the heat is transferred to the system in an isobaric process some amount of work is done. Along with the work, some change is also seen in the internal energy of the system. The other name of the isobaric process is the constant pressure process.
Here, Pi=Pf(ΔP=0){{P}_{i}}={{P}_{f}}(\Delta P=0)
Now, in the question it is given that work is done against the external pressure. So, the process is an irreversible process. Therefore the formula for work done is given as:
W=−PextΔVW=-{{P}_{ext}}\Delta V…………. (1)
Where, W= Work done in isobaric irreversible process
−Pext-{{P}_{ext}}= external pressure (here it is constant)
ΔV\Delta V= Change in the volume
Where, ΔV=(V2−V1)\Delta V=({{V}_{2}}-{{V}_{1}})
In the question value of constant external pressure is given as 1 atm and the value of V1{{V}_{1}} is given as 3dm33d{{m}^{3}}and V2{{V}_{2}} is given as 13 lire= 13dm3(1l=1dm3)13d{{m}^{3}}(1l=1d{{m}^{3}})
Now put all the values in equation (1), we will get:
W=−1 atm(13−3)dm3W=-1\,atm(13-3)d{{m}^{3}}
∴W=−10dm3\therefore W=-10d{{m}^{3}}

Thus, the correct option will be (A).

Note: Work done for isobaric process in reversible condition is given as W=−nRTln⁡[V2V1]W=-nRT\ln \left[ \dfrac{{{V}_{2}}}{{{V}_{1}}} \right]. In the question change the units wherever required. Isobaric and isochoric processes are different because in isobaric process pressure is kept constant while in isochoric process volume is kept constant.