# Work done in Isothermal Irreversible Process

## Trending Questions

**Q.**

2 litres of an ideal gas at a pressure of 10atm expands isothermally into vacuum until its total volume is 10 litres. How much work is done? How much heat is absorbed? Consider the same expansion but this time the constant external pressure of 1 ATM?

**Q.**

Five moles of an ideal gas at $1\mathrm{bar}$ and $298\mathrm{K}$ is expanded into the vacuum to double the volume. The work done is

Zero

${\mathrm{C}}_{\mathrm{v}}[{\mathrm{T}}_{2}-{\mathrm{T}}_{1}]$

$-\mathrm{RT}({\mathrm{V}}_{2}-{\mathrm{V}}_{1})$

$-\mathrm{RT}\mathrm{ln}({\mathrm{V}}_{2}/{\mathrm{V}}_{1})$

**Q.**The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm=101.32 J)

- −600 J
- −608 J
- +304 J
- −304 J

**Q.**Calculate the work done during isothermal expansion of 1 mol of an ideal gas from 10 atm to 1 atm at 300 K.

- - 1382 cal
- - 1382 J
- - 13.82 cal
- - 1.382 cal

**Q.**

The pressure-volume work for an ideal gas can be calculated by using the expression W=∫VfVipexdV. The work can also be calculated from the pV-plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf. Choose the correct option.

(a) W (reversible) = W (irreversible)

(b) W (reversible) < W (irreversible)

(c) W (reversible) = W (irreversible)

(d) W (reversible) = W (irreversible) +pex.ΔV

**Q.**Based on the first law of thermodynamics, which one of the following is correct?

- For an isothermal process, q=+w
- For an isochoric process, ΔU=−q
- For an adiabatic process, ΔU=−w
- For a cyclic process, q=−w

**Q.**Calculate the amount of work done by 2 mole of an ideal gas at 298k in reversible isothermal expansion from 10 liter to 20 liter.

-3435 J

3435 J

-3400 J

3400 J

**Q.**

One mole of an ideal gas expands reversibly at a constant temperature of $300\mathrm{K}$ from an initial volume of $10\mathrm{L}$ to a final volume of $20\mathrm{L}$. What will be the work done in expanding the gas?

$750\mathrm{J}$

$1500\mathrm{J}$

$1728\mathrm{J}$

$3456\mathrm{J}$

**Q.**

One mole of an ideal gas at 25∘C expands in volume from 1.0 L to 4.0 L at constant temperature. What work (in J) is done if the gas expands against vacuum ?

-4.0 X 10

^{2}4.0 X 10

^{2}J0 J

- 3.0 X10

^{2}

**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

- T = 174.8 K, V = 64 L
- T = 165.4 K, V = 78.8 L
- T = 153 K, V = 57 L
- T = 161.2 K, V = 68.3 L

**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.**In a thermodynamic process a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is reached and its pressure is then found to be 0.75 of its initial pressure. Find the adiabatic exponent of the gas

**Q.**A thermodynamic system goes from states (i) P1, V to 2P1, V and

(ii) P, V1 to P, 2V1

The work done in the two cases respectively is:

- Zero, Zero
- Zero, −PV1
- −PV1, Zero
- −PV1, −P1V1

**Q.**Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre 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.**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
- −810.4 J
- −212.8 J

**Q.**Assertion :The amount of work done in the isothermal expansion is greater than work done in the adiabatic system for same final volume. Reason: In the adiabatic expansion of a gas temperature and pressure both decrease due to decrease in internal energy of the system.

- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect but Reason is correct

**Q.**Find the work done when 5 moles of hydrogen contracts isothermally from 15 L to 5 L against a constant pressure of 1 atm at 250C.

- 685.3 cal
- −848.2 cal
- 422.5 cal
- −240.5 cal

**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.

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

**Q.**Work done on the gas, W, in an irreversible expansion is given by W=−Pext×dV, where Pext is the external pressure and dV is the change in volume of the gas (i.e., Vfinal−Vinitial). What is the value of work done on 1 mole of an ideal gas when it expands into vaccum?

- 2.27 J
- Infinite
- 0 J
- 8.314 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
- 300 J
- −100 J
- −600 J

**Q.**

5 moles of an ideal gas at 270C expands isothermally and reversibly from a volume of 6L to 60L. The work done in kJ is :

- -14.7
- -28.72
- + 28.72
- - 56.72

**Q.**A thermodynamic system goes from states (i) P1, V to 2P1, V and

(ii) P, V1 to P, 2V1

The work done in the two cases respectively is:

- Zero, Zero
- Zero, −PV1
- −PV1, Zero
- −PV1, −P1V1

**Q.**An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III The correct plot(s) representing the changes from state I to state III is(are)

(p:pressure, V:volume, T:temperature, H:enthalpy, S:entropy)

**Q.**Sample of air is saturated with the benzene( vapour pressure is equal 200 mmhgat 298 2Kelvin )298 Kelvin 750 mm HG pressure if it is isothermal comprrssed to one third of its initial volume the find final pressure of the system A 2250 torr B 2150 torr C 2050 torr D 1950 torr

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

**Q.**Assertion :Work done during free expansion of an ideal gas whether reversible or irreversible is positive. Reason: During free expansion, external pressure is always less than the pressure of the system.

- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Both Assertion and Reason are incorrect

**Q.**

One mole of an ideal gas at 25∘C expands in volume from 1.0 L to 4.0 L at constant temperature. What work (in J) is done if the gas expands against vacuum ?

−4.0 × 102

−3.0 × 102

4.0 × 102 J

0 J

**Q.**For an irreversible isothermal expansion of an ideal gas:

- ΔSsys=ΔSsurr
- ΔSsys=−ΔSsurr
- |ΔSsys|=|ΔSsurr|
- None of these

**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.**If w1, w2, w3 and w4 are work done in isothermal, adiabatic, isobaric and isochoric reversible expansion for an ideal gas, respectively, then:

- w3>w1
- w1>w2
- w2>w4
- w4>w2