1
Question
A system absorbs 300 cal of heat, its volume doubles and temperature rises from 273 to 298 k, the work done on the surrounding is 200 cal. ΔE for the above reaction is :
A system absorbs 300 cal of heat, its volume doubles and temperature rises from 273 to 298 k, the work done on the surrounding is 200 cal. ΔE for the above reaction is :
Open in App
Solution
The correct option is D 100 cal
As per the sign convention, the heat absorbed by the system has positive sign. Hence, q=+300cal
When work is done by the system on the surroundings, it has negative sign.
Thus w=−200cal.
The expression for the change in the internal energy is ΔE=q+w=+300cal−200cal=100cal
Thus, the net internal energy of the system increases by 100 cal.
As per the sign convention, the heat absorbed by the system has positive sign. Hence, q=+300cal
When work is done by the system on the surroundings, it has negative sign.
Thus w=−200cal.
The expression for the change in the internal energy is ΔE=q+w=+300cal−200cal=100cal
Thus, the net internal energy of the system increases by 100 cal.
Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
