Fixed mass of an ideal gas collected in a 24.64 litre sealed rigid vessel at 10 atm is heated from `-73^(@)C` to `27^(@)C` calculate change in gibbs free energy if entropy of gas is a function of temperature as `S=2+10^(-2)T(J//K)(1l atm=0.1kJ)`
A
`1231.5` J
B
`1281.5` J
C
`781.5` J
D
0
Text Solution
Verified by Experts
The correct Answer is:
C
At constant volume, `(P_(1))/(T_(1))=(P_(2))/(T_(2))` `rArr" "P_(2)=1xx(300)/(200)=(3)/(2)` and `V_(1)=24.63L` for single phase `because" "dG=Vdp-"S dT"` `DeltaG=V.DeltaP-int(2+10^(-2)T).dT` `=1231.5-200-(10^(-2)xx50,000)/(2)` = 781.5 J
Topper's Solved these Questions
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|6 Videos
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|2 Videos
One mole of an ideal gas for which C_(V)= 3//2R heated at a constant pressure of 1 atm from 25^(0)C "to" 100^(0)C What will be the amount of work done in the process?
One mole of an ideal gas for which C_(V)= 3//2R heated at a constant pressure of 1 atm from 25^(0)C "to" 100^(0)C What will be the amount of heat change at constant pressure?
An ideal gas expands from 500 cm^3 to 700 cm^3 against 1 atm pressure, by absorbing 2J of energy. Calculate change in internal energy of the ideal gas.
10 mole of an ideal gas is heated at constant pressure of one atmosphere from 27^(@)C to 127^(@)C . If C_(v, m) = 21.686 + 10^(-3) T , then Delta H for the process is x xx 10^(5) J1 . Then x is
An ideal gas at a pressure of 1 atm and temperature of 27^(@)C is compressed adiabatically until its pressure becomes 8 times the initial pressure , then final temperature is (gamma=(3)/(2))
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column