A 1.0 mol of ideal gas, initially at 10 atm and 300 K Is allowed to expand isothermally to 1.0 aim. as follows :
Inversible expansion against 1.0 atm
Determine the net work done by the system and describe the whole processes on p-V diagram.

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum untill its total volume is 10 litres. How much heat is absorbed & how much work is done in the expansion ?

When 1 mol of an ideal gas is compressed in a reversible isothermal process at T K, the pressure of the gas changes from 1 atm to 10 atm. In the process, if the work done by the gas be 5.744kJ, then T is-

One mole of an ideal gas at 300 K is expanded ‘ isothermally from an Initial volume of ( litre to 10 litre. What is the value of AE for this process ?

A 5 mol sample of an ideal gas is compressed isothermally and irreversible from 1.5 atm to 15 atm at 27^(@)C . Calculate the work done on the gas in calorie unit.

The pressure of 3 mol of an ideal gas is 10 atm at 27^(@)C . Calculate work done by the gas when it is expanded isothermally against an external pressure of 1 atm.

At 27^(@)C, 4.0 g of He is expanded reversibly and isothermally when the pressure drops from 10 to 1.0 atm. Calculate the work done in calorie, assuming an ideal behavior of the gas.

The initial state of 1 mol of an ideal gas is (P_(1),V_(1),T_(1)) . The gas is expanded by a reversible isothermal process and also by a reversible adiabatic process separately. If final volume of the gas is same in both the processes, and changes in internal energy in the isothermal and adiabatic processes are DeltaU_(1) and DeltaU_(2) respectively, then-

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vaccum until final volume is of 10 litres conducted reversibly. How much heat is absorbed ?

10 mol of an ideal gas is taken through an isothermal process in which the volume is compressed from 40 L to 30 L. If the temperature and the pressure of the gas are 0^(@)C and 1 atm respectively. Find the work done in the process. Given: R = 8.31 J cdot mol^(-1) cdot K^(-1) .