A sample of oxygen gas expands its volume from `3L` to `5L` against a constant pressure of `3` atm. If work done during expansion be used to heat `10 ` mole of water initially present at `290K` , its finally temperature will be `(` specific heat capacity of water `=4.18 J//k-g) :`

The work done during expansion is used to heat 10 mole of water of temperature of 290 K.

A gas expands from 3 dm^(3) to 5dm^(3) against a constant pressure of 3atm . The work done during the expansion if used to heat 10mol of water at temperature 290K . Find the final temperature of water, if the specific heat of water = 4.18g^(-1)K^(-1) .

A gas expands from 3 dm^(3) to 5dm^(3) anainst a constant pressure of 3atm . The work done during the expansion if used ti heat 10mol of water at temperature 290K . Find the final temperature of water, if the specific heat of water = 4.18g^(-1)K^(-1) .

A gas expands from 3 dm^(3) to 5 dm^(3) against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperature of water. Specific heat of water =4.184 J g^(-1)K^(-1)

A gas expands from 3 dm^(3) to 5 dm^(3) against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperature of water. Specific heat of water =4.184 J g^(-1)K^(-1)

A gas expands from 3 dm^3" to "5 dm^3 against a constant pressure of 3.0 atm. The work done during expansion is used to heat 10 moles of water at temperature 290 K. Calculate the final temperature of water. Specific heat of water = 4.184 J K^(-1) g^(-1) .

A gas expands from a volume of 3.0 dm^(3) to 5.0 dm^(3) against a constant external pressure of 3.0 atm. The work done during the expansion is used to heat 10.0 ml of water of temperature 290.0 K.Calculate the final temperature of water (specific heat of wate = 4.184 J g^(-1) K^(-1) ).

A gas expands in volume from 2 L to 5 L against a pressure of1atm at constant temperature. The work done by the gas will be