one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.o l to 2.0 l against a constant pressu
Answer (1)
7th Feb, 2019
If your complete question is:
One mole of an ideal gas at 300 K in thermal contact with the surroundings expands isothermally from1 L to2L against a constant pressure of 3 atm. In this process the change in the enthalapy of surroundings is? (change in S in J /K )1Latm.=101.3J
Ans: From First Law of Thermodynamics,
q(system)=del U- del W
Since its an isothermal reaction, del U = 0, del W = -P(ext.) x del V
so q(system) = 0-(-P(ext) x del V)
= 3 x (2-1)
= 3 L.atm
change in enthalphy of system = del S(system) = q/T = 3L.atm/300k = (3 x 101.3)J/300k
= 1.013 J/K
So, change in enthalphy of surrounding = del S(surrounding) = - del S(system) =
- 1.013 J/K
Comments (0)
