1
Question
Two moles of an ideal gas is heated at a constant pressure of one atmosphere from 27∘C to 127∘C. If Cv,m = 20 + 10−2 T JK−1 mol−1, then q and ΔU for the process are:
Two moles of an ideal gas is heated at a constant pressure of one atmosphere from 27∘C to 127∘C. If Cv,m = 20 + 10−2 T JK−1 mol−1, then q and ΔU for the process are:
Open in App
Solution
The correct option is A 6362.8 J, 4700 J
First thing to realise here is that this is an isobaric process. Hence, W=pdV=nRdT
W = −nRΔT = −2 × 8.314 × 100 = −1662.85
From the ideas learned in heat capacity, we know that dU=mCdT. On integration, you get:
ΔU = n∫ Cv,mdT = 2∫ (20 + 10−2T)dT
= 2 × 20 × (T2 − T1) + 2 × 10−2 × (T22 − T212) = 4700J
ΔU = WΔU = q − W
4700J = q − 1662.8
q = 6362.85
Tip: You can use R=25/3 to make quicker calcuations.
Remember this works only if the units are J/(molK)
6362.8 J, 4700 J
First thing to realise here is that this is an isobaric process. Hence, W=pdV=nRdT
W = −nRΔT = −2 × 8.314 × 100 = −1662.85
From the ideas learned in heat capacity, we know that dU=mCdT. On integration, you get:
ΔU = n∫ Cv,mdT = 2∫ (20 + 10−2T)dT
= 2 × 20 × (T2 − T1) + 2 × 10−2 × (T22 − T212) = 4700J
ΔU = WΔU = q − W
4700J = q − 1662.8
q = 6362.85
Tip: You can use R=25/3 to make quicker calcuations.
Remember this works only if the units are J/(molK)
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
