One mole of an ideal diatomic gas `(C_(v) =5 cal)` was transformed from initial `25^(@)` and `1 L` to the state when temperature is `100^(@)` C and volume `10` L. The entropy change of the process can be expressed as (`R=2` calories /mol/K)

One mole of an ideal diatomic gas (C_(v)=5cal) was transformed form initial 25^(@)C and 1L to the state when the temperature is 100^(@)C and volume 10L. The entropy change of the process can be express as: (R=2cal//mol//K)

One mole of an ideal diatomic gas (C_(v)=5cal) was transformed form initial 25^(@)C and 1L to the state when the temperature is 100^(@)C and volume 10L. The entropy change of the process can be express as: (R=2cal//mol//K)

One mole of an ideal diatomic gas (C_(v) =5 cal) was transformed from initial 25^(@) and 1 L to the state when temperature is 100^(@) C and volume 10 L. Then for this process(R= 2 "calories" //"mol"//K" )(take calories as unit of energy and kelvin for temp)

One mole of an ideal diatomic gas (C_(v)=5 cal K^(-1) "mole"^(-1)) chagre its state from state 1 (27^(@) C 1 L) to state 2 (127^(@) C, 10L) . For this process, which of the following is (are) correct? (Given : R=2 cal K^(-1) "mole"^(-1))

One mole of an ideal gas expands isothermally from 10 L to 100 L. Change in entropy of the process will be

The temperature and pressure of one mole of an ideal monoatomic gas changes from 25^(@) C and 1 atm pressure to 250^(@) C and 25 atm. Calculate the change in entropy for the process.

The temperature and pressure of one mole of an ideal monoatomic gas changes from 25^(@) C and 1 atm pressure to 250^(@) C and 25 atm. Calculate the change in entropy for the process.

When 5 moles of an ideal gas is expended at 350 K reversibly from 5 L to 40 L, then the entropy change for the process is (R = 2 cal mol^-1 K^-1 )