Calulate the work done (in cal.). When 1.0 mole of `N_(2)H_(4)` decomposes completely against a pressure of `1.0` atm at `27^(@)C` `(Given R=2 cal//mol//K)`
`N_(2)H_(4)(l)toNH_(3)(g)+N_(2)(g)`

Calculate the work done (in J) when 4.5 g of H_(2)O_(2) reacts against a pressure of 1.0 atm at 25^(@)C 2H_(2)O_(2)(l)rarr O_(2)(g) +2H_(2)O(l)

Calculate the work done (in J) when 4.5 g of H_(2)O_(2) reacts against a pressure of 1.0 atm at 25^(@)C 2H_(2)O_(2)(l)rarr O_(2)(g) +2H_(2)O(l)

Calculate the work done on the system in Joules when 2.0 moles of N_(2) reactes with 6.0 moles H_(2) , to form NH_(3) against a pressure of 1 .0 atm at 27^(@) C. N_(2)(g)+3H_(2)(g)to2NH_(3)(g) (Given : R= 8 J//mol K )

Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is (R = 2 cal/mol/K)

Work done by 0.1 mole of a gas at 27^(@)C to double its volume at constant pressure is (R=2" cal mol"^(-1).^(@)K^(-1))

Work done by 0.1 mole of a gas at 27^@ C to double its volume at constant pressure is (use R=2 cal/mol ""^(@) C)

2.8 g of N_(2) , 0.40 g of H_(2) and 6.4g of O_(2) are placed in a container of 1.0 L capacity at 27^(@)C . The total pressure in the container is :