JEE MAIN - Chemistry (2013 (Offline) - No. 1)
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant
temperature of 37.0oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be :
(R = 8.314 J/mol K) ( l n 7.5 = 2.01)
(R = 8.314 J/mol K) ( l n 7.5 = 2.01)
q = – 208 J, w = – 208 J
q = – 208 J, w = + 208 J
q = + 208 J, w = + 208 J
q = + 208 J, w = – 208 J
توضيح
By Ist
law of thermodynamics, q = $$\Delta $$E – W
At const T, ∆E = 0
q = – W
Heat absorbed = 208 J
$$ \therefore $$ q = +208 J
W = – 208 J
At const T, ∆E = 0
q = – W
Heat absorbed = 208 J
$$ \therefore $$ q = +208 J
W = – 208 J