# Work done in Isothermal Irreversible Process

## Trending Questions

**Q.**Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change ( in kJ) for the process is

0 kJ

4.8 kJ

−11.4 kJ

11.4 kJ

**Q.**The change in entropy of 2 moles of an ideal gas upon isothermal expansion at 243.6 K from 20 litre until the pressure becomes 1 atm, is

1.385 cal/K

− 1.2 cal/K

1.2 cal/K

2.77 cal/K

**Q.**36. An ideal gas expands isothermally from a volume V1 to V2 and then compressed to original volume V1 adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W. Then

**Q.**Calculate entropy change when 10 moles of an ideal gas expands reversibly and isothermally from an initial volume of 10 litre to 100 litre at 300K.

**Q.**10 litres of a monoatomic ideal gas at 0oC and 10 atm pressure is suddenly exposed to 1 atm pressure and the gas expands adiabatically against this constant pressure to maximum possible volume. The final temperature and volume of the gas respectively are

**Q.**Assertion :Statement-I : Entropy change in reversible adiabatic expansion of an ideal gas is zero.

Because Reason: Statement-II : The increase in entropy due to volume increase just compensate the decrease in entropy

due to fall in temperature.

- Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
- Statement-I is true, Statement-II is true ; Statement-II is not a correct explanation for Statement-I.
- Statement-I is true, Statement-II is false.
- Statement-I is false, Statement-II is true.

**Q.**Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is:

(Given that 1 L bar = 100 J)

- -30 J
- 5 kJ
- 25 J
- 30 J

**Q.**Match the following

List IList IIA. Reversible isothermal expansion(p)w=−RPext[T2P1−T1P2]P1P2B. Irreversible isothermal expansion (q)w=Rγ−1[T2−T1]C. Reversible adiabatic expansion(r)w=−Pext(V2−V1)D. Irreversible adiabatic expansion(s)w=−nRT log(V2V1)

- A - (r), B - (s). C-(q), D - (p)
- A - (s), B - (r). C-(p), D - (q)
- A - (s), B - (r). C-(q), D - (p)
- A - (p), B - (r). C-(q), D - (s)

**Q.**110.1 L of an ideal gas( Y=1.5) at 300 K is suddenly compressed to half of its original volume.Original pressure is 100 kPa.The gas is now expanded isothermally to achieve its original volume of 1 L.Calculate the work done by the gas?

**Q.**Two moles of an ideal gas is expanded isothermally and reversibly from 1 L to 10 L at 300 K. The internal energy change ( in kJ) for the process is

- 11.4 kJ
- -11.4 kJ
- 0 kJ
- 4.8 kJ

**Q.**One mole of a gas is expanded from (1L, 10 atm, 300 K) to (4 L, 5 atm, 300 K) against a constant external pressure of 1 atm. The heat capacity of gas is 50 J/oC.Then the work done during the process is:

(Take 1Latm≃100J)

- −300 J
- −600 J
- 300 J
- −100 J

**Q.**If a gas at 10 atm and 300 K expands against a constant external pressure of 2 atm from a volume of 10 litres to 20 litres, the amount of work done is:

Given: 1 L.atm = 101.3 J

- 17 L.atm
- 10 L.atm
- −2026 J
- 2026 J

**Q.**If a gas at 5 atm and 373 K expands against a constant external pressure of 1 atm from a volume of 2 litres to 10 litres, the amount of work done is:

(Given: 1L.atm=101.3 J)

- −402 J
- −525.3 J
- −212.8 J
- −810.4 J

**Q.**If one mole of an ideal gas (Cp, m=52R) is expanded isothermally at 300 K, until it's volume is tripled, then, change in entropy of gas is :

- ∞
- 52Rln3
- 0
- Rln3

**Q.**An ideal gas undergoing expansion in vaccum shows

1) deltaE= 0

2) W=0

3 q=0

Why? Give explaination.

**Q.**Match the following

List IList IIA. Reversible isothermal expansion(p)w=−RPext[T2P1−T1P2]P1P2B. Irreversible isothermal expansion (q)w=Rγ−1[T2−T1]C. Reversible adiabatic expansion(r)w=−Pext(V2−V1)D. Irreversible adiabatic expansion(s)w=−2.303 nRT log10V2/V1

- A - (s), B - (r). C-(q), D - (p)
- A - (s), B - (r). C-(p), D - (q)
- A - (r), B - (s). C-(q), D - (p)
- A - (p), B - (r). C-(q), D - (s)

**Q.**Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically. The magnitude of work done is

- More in the isothermal process
- More in the adiabatic process
- Equal in both processes
- Dependent on atomicity of gas

**Q.**a very good evening to

__RESPECTED EXPERTS__

A GAS EXPANDS ISOTHERMALLY from 10 dm3 to 20 dm3 at 27 degree celsius and work obtained is 4.620 kj. calculate the number of mols of the gas.

**Q.**calculate the work done in joule when 3 moles of an ideal gas expand isothermally and reversibly

from 10 atm to 1 atm. what will be the work done if the expansion is against constant pressure of 1atm.

**Q.**Calculate the heat involved in a reaction for the isothermal expansion of one mole of a ideal gas at 27o C from a volume of 50 L to 100 L.

- Q = 159 J
- Q = 2294 J
- Q = 1729 J
- Q = 2197 J

**Q.**Two moles of an ideal gas is expanded isothermally and reversibly from 1 L to 10 L at 300 K. The internal energy change ( in kJ) for the process is

- 11.4 kJ
- -11.4 kJ
- 0 kJ
- 4.8 kJ

**Q.**2 mole of a ideal gas at 2 bar and 27 deg celcius expands isothermally against a constant pressure of 1 bar. the work done by the gas is equal to

**Q.**5 moles of ideal gas at temp. T is compressed isothermally from 12 atm to 24 atm. Calculate the value of 10r,

where, r=work done in reversible processwork done in irreversible process

(Given : ln2=0.7)

**Q.**

3 moles of ideal gas is heated as constant pressure from 27 degree cantegrate to 127 degree centigrade.

(a) Find the workdone for the expansion of gas?

(b) If the gas were expanded isothermally in a irreversible manner at 27 degree centigrade from 1 atmosphere to 0.7 atmosphere. Calculate the workdone.

**Q.**Calculate the work done when 5 moles of oxygen gas contracts isothermally from 85 L to 15 L against a constant pressure of 2 atm at 250C.

- 70 L atm
- 14.0 L atm
- 140 L atm
- 7.0 L atm