Question

A gas expands in volume from 2L to 5L against a pressure of 1 atm at constant termperature. The work done by the gas will be (a) 3 J(b) -303.9 J(c) -303.9 L atm(d) 303.9 L atm

Answer 1

Answer:

b

Explanation:

P∆v

then change into joule

Answer 2

Answer:

(b) -303.9 J

Explanation:

Given,

The gas is expanding,

initial volume of gas = 2 L

final volume of gas = 5 L

change in volume v = final volume - initial volume

                                 =5 L - 2 L

                                 = 3 L

pressure = 1 atm

Since, the pressure is constant

Hence, work done = pressure x change in volume

                                = 1 atm x 3 L

                                 = 3 L atm

Since, 1 L atm= 101.325 J

 so, 3 L atm= 3 x 101.325 J

                    = 303.9 j

Since, the work done by the system so, the work done should be negative and the work done will be -303.9 J.