ThermodynamicsMedium
Question
When water is boiled at 2 atm pressure, the letant heat of vaprization is 2. × 106 J/ kg and the boiling point is 120oC. At 2 atm pressure, 1 atm pressure, 1 kg of water has volume of 10-3 m3 and 1 kg of steam has a volume of 0.824 m3. The increase in internal energy of 1 kg of water when it is converted into steam at 2atm pressure and 120oC is [1 atm pressure = 1.013 × 105N/m2]
Options
A.2.033 J
B.2.033 × 106 J
C.0.167 × 106 J
D.2.267 × 106 J
Solution
Total heat given to convert water into steam at 120oC is Q = mL 1 × 2.2 × 106 = 2.2 × 106 J
The work done by the system against the surrounding is
PᐃV = 2 × 1.013 × 105(0.824 × 0.001) = 0.167 × 106 J ∴ ᐃU = Q - W = 2.033 × 106 J
The work done by the system against the surrounding is
PᐃV = 2 × 1.013 × 105(0.824 × 0.001) = 0.167 × 106 J ∴ ᐃU = Q - W = 2.033 × 106 J
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
The volume of a poly-atomic gas compressed adiabatically to of the original volume. If the original pressure of the gas ...The molar heat capacity at constant presure of nitrogen gas at STP is nearly 3.5 R. Now when the temperature is increase...dU + δW = 0 is valid for :-...The value of internal energy in an adiabatic process :-...-If DA and BC are adiabatic curves and AB and CD are isothermal curves then : -If DA and BC are adiabatic curves and AB ...