Solve this :
Q. When a sample of gas expands from 4.0 L to 12.0 L against a constant pressure of 0.30 atm, the work involved is
1) 243.19 J 2) -243.19 J
3) 234.19 J 4) -234.19 J
Dear student,
Given
V1 = 4 lit
V2 = 12 lit
P=0.30atm
Formula used :
W=−P∆V =−2×(20−10) =−20 l atm W = - PV
= - 0.30 ( V2- V1 )
= - 0.30 8
= - 2.4 lit atm
1 lit atm = 101.33 J
Therefore,
W = -243.19 J
so option 2 is correct
Regards
W=−P∆V =−2×(20−10) =−20 l atm
Given
V1 = 4 lit
V2 = 12 lit
P=0.30atm
Formula used :
W=−P∆V =−2×(20−10) =−20 l atm W = - PV
= - 0.30 ( V2- V1 )
= - 0.30 8
= - 2.4 lit atm
1 lit atm = 101.33 J
Therefore,
W = -243.19 J
so option 2 is correct
Regards