The amount of work done when a gas is expanded from a volume of 5 L to 10 L against a constant external pressure of 2 atm is used to heat 15 g of water present at a temperature of 300 K. Calculate the final temperature of water. (Given : molar heat capacity of water `= 75.6 J. K^(-1). " mol"^(-1)`).

Explanation

We know, `w= - P_(ex)(V_(2)-V_(1))`
Given that `P_(ex)=2 " atm" , V_(1)=5 L and V_(2)=10 L`.
` therefore w= - 2(10-5)` L. atm = - 10 L atm= -1012.66 J
15 g of water `=(15)/(18)` mol of water. Let the temperature of water becomes TK when it is heated.
So, the amount of heat supplied to water
`= n C(T-300)= (15)/(18)xx75.6(T-300) J`
`[ therefore n= (15)/(18) " mol and " C= 75.6 J. K^(-1) " mol"^(-1)]`
It is told that work done in the expansion of gas has been used to heat the water.
Therefore, `1012.66= (15)/(18)Xx75.6(T-300)`
` therefore T= 316.07 K`
Hence, the final temperature is 316.07 K.