Thermodynamics | Questions based on Work done
Questions
5 moles of an ideal gas expands isothermally and reversibly from 1 L to 10 L at 27oC. 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.
\[Given\ values: T = 300\ K,\ V_1 = 1L,\ V_2 = 10L\]
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} {nRT \over V} dV\]
\[W = - nRT \int\limits_{V_1}^{V_2} {1 \over V} dV\]
\[W = - nRT\ ln {V_2 \over V_1}\]
\[W = -5 \times 8.314 \times 300 \ln {10 \over 1}\]
\[W = -28720.713\ J\]
10 moles of an ideal gas expands isothermally and reversibly from pressure 12 atm to 2 atm at 127oC. 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.
\[Given\ values: T = 400\ K,\ P_1 = 12\ atm,\ P_2 = 2\ atm\]
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} {nRT \over V} dV\]
\[W = - nRT \int\limits_{V_1}^{V_2} {1 \over V} dV\]
\[W = - nRT\ ln {V_2 \over V_1}\]
\[Since,\ V ∝ {1 \over P}\]
\[W = - nRT\ ln {P_1 \over P_2}\]
\[W = -10 \times 8.314 \times 400 \ln {12 \over 2}\]
\[W = -\ 59586.75\ J\]
3 moles of an ideal gas expands irreversibly at 27oC 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:
\[Given\ values: T = 300\ K,\ P_1 = 20\ atm,\ P_2 = 10\ atm\]
\[W = - P_{ext}(V_2 - V_1)\]
\[W = - P_{ext}({nRT \over P_2} - {nRT \over P_1})\]
\[W = - P_{ext} nRT ({1 \over P_1} - {1 \over P_2})\]
\[W = -10 \times 3 \times 0.0821 \times 300 \times ({1 \over 10} - {1 \over 20})\]
\[W = -36.945\ atm\ L\]
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:
\[Given\ values: V_1 = 3\ L,\ V_2 = 1\ L\]
\[W = - P_{ext}(V_2 - V_1)\]
\[W = - 10 (3 - 1) = -20\ atm\ L\]
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 127oC and 27oC, calculate work done.
Since this is a reversible process, work done can be calculated as:
\[Given\ values: T_1 = 400\ K,\ T_2 = 300\ K\]
Since, this is an isobaric reversible process, so let Pgas = Pext = P.
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - P_{ext} \int\limits_{V_1}^{V_2} dV\]
\[W = -P_{ext} (V_2 - V_1)\]
\[W = -P_{ext} ({nRT_2 \over P_{gas}} - {nRT_1 \over P_{gas}})\]
\[W = -nRP ({T_2 \over P} - {T_1 \over P})\]
\[W = -nR (T_2 - T_1)\]
\[W = -5 \times 8.314 \times (300-400)\]
\[W = -5 \times 8.314 \times (-100)\]
\[W = 4157\ J\]
Find the work done when one mole of an ideal gas is expanded reversibly and isothermally from pressure 5 atm to 1 atm at 27oC.
Since this is a reversible process, work done can be calculated as:
\[Given\ values: T = 300\ K,\ P_1 = 5\ atm,\ P_2 = 1\ atm\]
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} {nRT \over V} dV\]
\[W = - nRT \int\limits_{V_1}^{V_2} {1 \over V} dV\]
\[W = - nRT\ ln {V_2 \over V_1}\]
\[W = - nRT\ ln {P_1 \over P_2}\]
\[W = -1 \times 8.314 \times 300 \ln {5 \over 1}\]
\[W = - 4014.26\ J\]
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:
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} ({αV + β}) dV\]
\[W = - {\int\limits_{V_1}^{V_2} (αV)dV } - β{\int\limits_{V_1}^{V_2}dV}\]
\[W = - {α \over 2}({V_2}^2 - {V_1}^2) - β(V_2 - V_1)\]
5 moles of a real gas following the equation P = αV2 is expanded from 1 L to 5 L isothermally and reversibly at 27oC. Calculate the work done by the gas.
Since this is a reversible process, work done can be calculated as:
\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} ({αV^2}) dV\]
\[W = - {α(V_2^3 - V_1^3) \over 3}\]
\[W = - {α(5^3 - 1^3) \over 3} = -{124α \over 3}\]
5 moles of a real gas undergoing isothermal irreversible expansion from volume 5 L to 10 L at 27oC 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:
\[W = - P_{ext} (V_2 - V_1)\]
\[W = -10 (10 - 5)\]
\[W = -50 atm\ L\]
\[W = -50 \times 101.3\ J\]
\[W = 5065\ J\]
5 moles of a real gas undergoing isothermal irreversible expansion from pressure 20 atm to 10 atm at 27oC under external pressure of 10 atm. Calculate work done if P(V-0.5n) = nRT.
This is an irreversible process, so work done = -Pext(V2 - V1).
\[Given:\ P(V - 0.5n) = nRT\]
\[V - 0.5n = {nRT \over P}\]
\[V = {nRT \over P} + 0.5n\]
\[Now,\ work\ done\ is\ given\ by:\]
\[W = - P_{ext} ({nRT \over P_2} + 0.5n - {nRT \over P_1} - 0.5n)\]
\[W = - nRTP_{ext} ({1 \over P_2} - {1 \over P_1})\]
\[W = -5 \times 0.0821 \times 300 \times 10 ({1 \over 10} - {1 \over 20})\]
\[W = -61.575\ atm\ L\]
\[W = -61.575 \times 101.3\ J\]
\[W = -6237.5475\ J\]