Thermodynamics | Questions based on Work done#

Questions#

5 moles of an ideal gas expands isothermally and reversibly from 1 L to 10 L at 27^oC. Calculate work done by the gas.

Since this is an isothermal reversible process, we will use integral to calculate total work done by the gas.

G i v e n v a l u e s : T = 300 K, V_{1} = 1 L, V_{2} = 10 L

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - \int_{V_{1}}^{V_{2}} P_{g a s} d V

W = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V

W = - n R T \int_{V_{1}}^{V_{2}} \frac{1}{V} d V

W = - n R T l n \frac{V_{2}}{V_{1}}

W = - 5 \times 8.314 \times 300 \ln \frac{10}{1}

W = - 28720.713 J

10 moles of an ideal gas expands isothermally and reversibly from pressure 12 atm to 2 atm at 127^oC. Calculate work done by the gas.

Since this is an isothermal reversible process, we will use integral to calculate total work done by the gas.

G i v e n v a l u e s : T = 400 K, P_{1} = 12 a t m, P_{2} = 2 a t m

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - \int_{V_{1}}^{V_{2}} P_{g a s} d V

W = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V

W = - n R T \int_{V_{1}}^{V_{2}} \frac{1}{V} d V

W = - n R T l n \frac{V_{2}}{V_{1}}

S i n c e, V \propto \frac{1}{P}

W = - n R T l n \frac{P_{1}}{P_{2}}

W = - 10 \times 8.314 \times 400 \ln \frac{12}{2}

W = - 59586.75 J

3 moles of an ideal gas expands irreversibly at 27^oC at an external pressure of 10 atm. If pressure of gas decreases from 20 atm to 10 atm, calculate work done by the gas.

Since this is an irreversible process, work done can be calculated as:

G i v e n v a l u e s : T = 300 K, P_{1} = 20 a t m, P_{2} = 10 a t m

W = - P_{e x t} (V_{2} - V_{1})

W = - P_{e x t} (\frac{n R T}{P_{2}} - \frac{n R T}{P_{1}})

W = - P_{e x t} n R T (\frac{1}{P_{1}} - \frac{1}{P_{2}})

W = - 10 \times 3 \times 0.0821 \times 300 \times (\frac{1}{10} - \frac{1}{20})

W = - 36.945 a t m L

Since, 1 atm L = 101.3 J, we can write:

W = - 36.945 \times 101.3 J = - 3742.5285 J

Fixed mass of helium gas expands irreversibly at constant temperature under an external pressure of 10 atm. If volume is increased from 1L to 3L, calculate work done.

Since this is irreversible process, work done can be calculated as:

G i v e n v a l u e s : V_{1} = 3 L, V_{2} = 1 L

W = - P_{e x t} (V_{2} - V_{1})

W = - 10 (3 - 1) = - 20 a t m L

Since, 1 atm L = 101.3 J, we can write:

W = - 20 \times 101.3 J = - 2026 J

5 moles of an ideal gas undergoes isobaric reversible process. If the initial and final temperatures are 127^oC and 27^oC, calculate work done.

Since this is a reversible process, work done can be calculated as:

G i v e n v a l u e s : T_{1} = 400 K, T_{2} = 300 K

Since, this is an isobaric reversible process, so let P_gas = P_ext = P.

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - P_{e x t} \int_{V_{1}}^{V_{2}} d V

W = - P_{e x t} (V_{2} - V_{1})

W = - P_{e x t} (\frac{n R T_{2}}{P_{g a s}} - \frac{n R T_{1}}{P_{g a s}})

W = - n R P (\frac{T_{2}}{P} - \frac{T_{1}}{P})

W = - n R (T_{2} - T_{1})

W = - 5 \times 8.314 \times (300 - 400)

W = - 5 \times 8.314 \times (- 100)

W = 4157 J

Find the work done when one mole of an ideal gas is expanded reversibly and isothermally from pressure 5 atm to 1 atm at 27_oC.

Since this is a reversible process, work done can be calculated as:

G i v e n v a l u e s : T = 300 K, P_{1} = 5 a t m, P_{2} = 1 a t m

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - \int_{V_{1}}^{V_{2}} P_{g a s} d V

W = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V

W = - n R T \int_{V_{1}}^{V_{2}} \frac{1}{V} d V

W = - n R T l n \frac{V_{2}}{V_{1}}

W = - n R T l n \frac{P_{1}}{P_{2}}

W = - 1 \times 8.314 \times 300 \ln \frac{5}{1}

W = - 4014.26 J

For a reversible process on a real gas, the volume expands from V₁ to V₂ at pressure P = αV+β. Calculate the work done.

Since this is a reversible process, work done can be calculated as:

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - \int_{V_{1}}^{V_{2}} P_{g a s} d V

W = - \int_{V_{1}}^{V_{2}} (α V + β) d V

W = - \int_{V_{1}}^{V_{2}} (α V) d V - β \int_{V_{1}}^{V_{2}} d V

W = - \frac{α}{2} ({V_{2}}^{2} - {V_{1}}^{2}) - β (V_{2} - V_{1})

5 moles of a real gas following the equation P = αV² is expanded from 1 L to 5 L isothermally and reversibly at 27^oC. Calculate the work done by the gas.

Since this is a reversible process, work done can be calculated as:

W = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

W = - \int_{V_{1}}^{V_{2}} P_{g a s} d V

W = - \int_{V_{1}}^{V_{2}} (α V^{2}) d V

W = - \frac{α (V_{2}^{3} - V_{1}^{3})}{3}

W = - \frac{α (5^{3} - 1^{3})}{3} = - \frac{124 α}{3}

5 moles of a real gas undergoing isothermal irreversible expansion from volume 5 L to 10 L at 27^oC under external pressure of 10 atm. Calculate work done if P(V-0.5n) = nRT.

Since this is an irreversible process, work done can be calculated as:

W = - P_{e x t} (V_{2} - V_{1})

W = - 10 (10 - 5)

W = - 50 a t m L

W = - 50 \times 101.3 J

W = 5065 J

5 moles of a real gas undergoing isothermal irreversible expansion from pressure 20 atm to 10 atm at 27^oC under external pressure of 10 atm. Calculate work done if P(V-0.5n) = nRT.

This is an irreversible process, so work done = -P_ext(V₂ - V₁).

G i v e n : P (V - 0.5 n) = n R T

V - 0.5 n = \frac{n R T}{P}

V = \frac{n R T}{P} + 0.5 n

N o w, w o r k d o n e i s g i v e n b y :

W = - P_{e x t} (\frac{n R T}{P_{2}} + 0.5 n - \frac{n R T}{P_{1}} - 0.5 n)

W = - n R T P_{e x t} (\frac{1}{P_{2}} - \frac{1}{P_{1}})

W = - 5 \times 0.0821 \times 300 \times 10 (\frac{1}{10} - \frac{1}{20})

W = - 61.575 a t m L

W = - 61.575 \times 101.3 J

W = - 6237.5475 J