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ChemistryEdu Logo Thermodynamics | Questions based on first Law of Thermodynamics#

Questions#

For an isochoric reversible process, calculate the following if Cv = 3R/2 for the gas present in the system and 5 moles of gas is heated from 27oC to 127oC.
(a) Work done (W)
(b) Heat supplied (q)
(c) Change in internal energy (ΔU)

(a) Since this is an isochoric process, work done will be zero.

(c) Change in internal energy (ΔU) is given by:

\[ΔU = \int\limits_{T_1}^{T_2} nC_vdT\]
\[ΔU = nC_v(T_2 - T_1)\]
\[ΔU = 5 \times {3R \over 2} \times (400 -300)\]
\[ΔU = 750R\]
\[ΔU = 750 \times 8.314\]
\[ΔU = 6235.5 J\]

(b) Heat supplied (q) is calculated by first law of thermodynamics:

\[ΔU = q + W\]
\[6235.5 = q + 0\]
\[q = 6235.5\ J\]

Five moles of an ideal gas at 300 K is expanded isothermally from an initial pressure of 4 atm to a final pressure of 1 atm against a constant external pressure of 1 atm. Calculate q, W, ΔU and ΔH. Calculate the corresponding value of all if this process is carried out reversibly.

(a) If the process is carried out irreversibly, let's calculate q, W, ΔU and ΔH.

  • Work done can be calculated as:
\[W = -P_{ext} (V_2 - V_1)\]
\[W = -P_{ext} ({nRT \over P_2} - {nRT \over P_1})\]
\[W = - nRT P_{ext}({1 \over P_2} - {1 \over P_1})\]
\[W = - 5 \times 0.0821 \times 300 \times 1 \times ({1 \over 1} - {1 \over 4})\]
\[W = -92.3625\ atm\ L\]
\[W = -92.3625 \times 101.3\ J = -9356.32\ J\]
  • Change in internal energy, ΔU is given by:
\[ΔU = \int\limits_{T_1}^{T_2}nC_vdT\]
\[ΔU = nC_v(T_2 - T_1)\]

Since, this is an isothermal process, so T2 = T1.

\[Hence,\ ΔU = 0\]
  • Change in enthalpy, ΔH is given by:
\[ΔH = \int\limits_{T_1}^{T_2}nC_pdT\]
\[ΔH = nC_p(T_2 - T_1)\]

Since, this is an isothermal process, so T2 = T1.

\[Hence,\ ΔH = 0\]
  • Heat supplied, q is calculated from first law of thermodynamics:
\[ΔU = q + W\]
\[0 = q + (-9356.32)\]
\[q = 9356.32\ J\]

(b) If the process is carried out reversibly, let's calculate q, W, ΔU and ΔH.

  • Work done is calculated as:
\[W = -\int\limits_{V_1}^{V_2}P_{ext}dV\]
\[W = -\int\limits_{V_1}^{V_2}P_{gas}dV\]
\[W = -\int\limits_{V_1}^{V_2}{nRT \over V}dV\]
\[W = -nRT\ ln{V_2 \over V_1}\]
\[W = -nRT\ ln{P_1 \over P_2}\]
\[W = -5 \times 8.314 \times 300 \times ln{4 \over 1}\]
\[W = -17288.48\ J\]

  • Since, the process is isothermal, so ΔU = 0 and ΔH = 0.

  • Heat supplied, q can be calculated from first law of thermodynamics:

\[ΔU = q + W\]
\[0 = q + (-17288.48)\]
\[q = 17288.48\ J\]

One mole of an ideal gas is (Cv = 20 J K-1mol-1) is allowed to expand reversibly and adiabatically from a temperature of 27oC. If the work done by the gas in the process is 3 KJ, what is the final temperature of the gas?

We know that for an adiabatic process, heat supplied (q) = 0.

By first law of thermodynamics:

\[ΔU = q + W\]
\[nC_v(T_2 - T_1) = 0 + W\]
\[nC_v(T_2 - T_1) = W\]
\[1 \times 20 (T_2 - 300) = - 3 \times 1000\]
\[T_2 - 300 = -150\]
\[T_2 = 150\ K\]

What is the change in internal energy when a gas contracts irreversibly from 377 mL to 177 mL under a constant pressure of 1520 torr while at the same time being cooled by removing 124 J of heat? (Take 1 atm = 760 torr and 1 atm L = 100 J)

The given process is irreversible. Heat is removed from the system, so its value will be negative. q = -124 J.

\[P_{ext} = 1520\ torr = {1520 \over 720}\ atm = 2\ atm\]
\[W = -P_{ext} (V_2 - V_1)\]
\[W = -2 (0.177 - 0.377)\]
\[W = 0.4\ atm\ L\]
\[W = 0.4 \times 100\ J = 40\ J\]

By first law of thermodynamics:

\[ΔU = q + W\]
\[ΔU = -124 + 40\]
\[ΔU = -84\ J\]

A real gas is subjected to an adiabatic process from (2 bar, 40 L, 300 K) to (4 bar, 40 L, 300 K) against a constant pressure of 4 bar. Calculate the enthalpy change for the process. (Take 1 bar L = 100 J)

Work done is given by:

\[W = -P_{ext} (V_2 - V_1)\]
\[W = -4 (30 - 40)\]
\[W = 40\ bar\ L\]

Enthalpy is given by:

\[H = U + PV\]
\[ΔH = ΔU + Δ(PV)\]
\[ΔH = ΔU + (P_2V_2 - P_1V_1)\]
\[ΔH = 40 + (4 \times 30 - 2 \times 40)\]
\[ΔH = 40 + (120 - 80)\]
\[ΔH = 80\ bar\ L\]
\[ΔH = 80 \times 100 = 8000\ J\]

Two moles of an ideal diatomic gas (Cv = 5R/2) at 300 K and 5 atm pressure is expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, W, ΔH and ΔU.

\[Poisson's\ ratio,\ γ = {C_p \over C_v}\]
\[γ = {C_v + R \over C_v}\]
\[γ = {{7R \over 2} \over {5R \over 2}} = {7 \over 5}\]
  • Equation of state for an adiabatic process is:
\[PV^γ = constant\]
\[P (T/P)^γ = constant\]
\[P^{1-γ} T^γ = constant\]
\[P_1^{1-γ} T_1^γ = P_2^{1-γ} T_2^γ\]
\[(5)^{1-7/5} \times (300)^{7/5} = (2)^{1-7/5} \times (T_2)^{7/5}\]
\[(5)^{-2/5} \times (300)^{7/5} = (2)^{-2/5} \times (T_2)^{7/5}\]
\[(T_2)^{7/5} = {(5)^{-2/5} \times (300)^{7/5} \over (2)^{-2/5}}\]
\[(T_2)^{7/5} = 2036.075\]
\[T_2 = 230.9\ K\]
  • Work done can be calculated as:
\[W = -P_{ext} (V_2 - V_1)\]
\[W = -P_{ext} ({nRT_2 \over P_2} - {nRT_1 \over P_1})\]
\[W = -nRP_{ext}({T_2 \over P_2} - {T_1 \over P_1})\]
\[W = -2 \times 0.0821 \times 1 ({230.9 \over 2} - {300 \over 5})\]
\[W = -28.81\ atm\ L\]
\[W = -28.81 \times 101.3\ J = -2918.34\ J\]
  • Change in internal energy for ideal gas is given by:
\[ΔU = nC_v(T_2 - T_1)\]
\[ΔU = 2 \times {5R \over 2} (230.9 - 300)\]
\[ΔU = -2872.487\ J\]
  • Heat supplied, q is calculated from first law of thermodynamics:
\[ΔU = q + W\]
\[-2872.487 = q - 2918.34\]
\[q = -2872.487 + 2918.34 = 45.853\ J\]
  • Change in enthalpy for ideal gas is given by:
\[ΔU = nC_p(T_2 - T_1)\]
\[ΔU = 2 \times {7R \over 2} (230.9 - 300)\]
\[ΔU = -4021.4818\ J\]

One mole of CO2 gas at 300 K is expanded under reversible adiabatic condition such that its volume becomes 27 times.
(a) What is the final temperature?
(b) Calculate work done.
Given: γ = 4/3, Cv=25.08 J mol-1K-1 for CO2

(a) Let initial volume, V1 be V and final volume, V2 be 27V.

The equation of state of an adiabatic process is:

\[PV^{γ} = constant\]
\[or,\ TV^{γ-1} = constant\]
\[T_1V_1^{γ-1} = T_2V_2^{γ-1}\]
\[300(V)^{4/3 - 1} = T_2 (27V)^{4/3 - 1}\]
\[300 (V)^{1/3} = T_2 (27V)^{1/3}\]
\[T_2 = 300 ({V \over 27V})^{1/3}\]
\[T_2 = 100\ K\]

Hence, final temperature is 100 K.

(b) By using first law of thermodynamics, we can calculate work done:

\[ΔU = q + W\]

Since, this is an adiabatic process, q = 0.

\[W = ΔU\]
\[W = nC_v(T_2 - T_1)\]

Since, Cv = R/(γ-1), we can write:

\[W = {nR (T_2 - T_1) \over γ-1}\]
\[W = {1 \times 8.314 (100 - 300) \over (4/3 - 1)}\]
\[W = -4988.4\ J\]

One mole of an ideal monoatomic gas is expanded irreversibly against an external pressure of 2 bar in two steps:
State a(8 bar, 4 L, 300 K) ----> State b(2 bar, 16 L, 300 K) ----> State c(1 bar, 32 L, 300 K)
Calculate total heat absorbed by the gas in this process.

Work done in path ab is given by:

\[W_{ab} = -P_{ext}(V_2 - V_1)\]
\[W_{ab} = -2 (16 - 4) = -24 bar\ L\]
\[W_{ab} = -24 \times 100\ J = -2400\ J\]

Work done in path bc is given by:

\[W_{bc} = -P_{ext}(V_2 - V_1)\]
\[W_{bc} = -2 (32 - 16) = -32 bar\ L\]
\[W_{bc} = -32 \times 100\ J = -3200\ J\]

Since, work done is a path function, we can calculate total work done as:

\[W = W_{ab} + W_{bc}\]
\[W = -2400 + (-3200) = -5600\ J\]

Using first law of thermodynamics, we can write:

\[ΔU = q + W\]
\[nC_v(T_3 - T_1) = q - 5600\]
\[nC_v(300 - 300) = q - 5600\]
\[0 = q - 5600\]
\[q = 5600\ J\]

Consider the given reaction at 300 K: H2(g) + Cl2(g) → 2HCl(g)
ΔH for this reaction is -185 KJ/mol. Calculate ΔU if 3 moles of H2 completely react with 3 moles of Cl2 to form HCl.

For a gaseous phase chemical reaction, we know that:

\[ΔH = ΔU + (Δn_g)RT\]

For the given reaction: Δng = 0.

\[So,\ ΔU = ΔH\]

Since, ΔH for 1 mole = -185 KJ, so ΔH for 3 moles = -185 * 3 = -555 KJ

\[ΔU = -555\ KJ\]

An ideal gas is expanded reversibly so that the amount of heat transferred to the gas is equal to the decrease in its internal energy. Derive the equation of this process in terms of variables T and V.

The amount of heat transferred to the gas is equal to the decrease in its internal energy means dq = -dU.

Using first law of thermodynamics:

\[ dU = dq + dW \]
\[ dU = -dU + dW \]
\[ 2dU = dW \]
\[2nC_vdT = - P_{ext}dV\]
\[2nC_vdT = - P_{gas}dV\]
\[2nC_vdT = - {nRT \over V}dV\]
\[Since,\ C_v = {R \over (γ-1)},\ we\ can\ write:\]
\[2n{R \over (γ-1)}dT = - {nRT \over V}dV\]
\[ {2 \over (γ-1)} {1 \over T}dT = - {1 \over V}dV\]
\[Integrating\ both\ sides:\]
\[ {2 \over (γ-1)} \int\limits_{T_1}^{T_2}{1 \over T}dT = - \int\limits_{V_1}^{V_2}{1 \over V}dV \]
\[ {2 \over (γ-1)}\ ln {T_2 \over T_1} = ln{V_2 \over V_1}\]
\[ ln ({T_2 \over T_1})^{2/(γ-1)} = ln{V_2 \over V_1} \]
\[({T_2 \over T_1})^{2/(γ-1)} = {V_2 \over V_1}\]
\[({T_2 \over T_1})^2 = ({V_2 \over V_1})^{(γ-1)}\]
\[({T_2 \over T_1})^2 = ({V_1 \over V_2})^{(1-γ)}\]
\[({T_2 \over T_1}) = ({V_1 \over V_2})^{(1-γ)/2}\]
\[T_1 V_1^{(1-γ)/2} = T_2 V_2^{(1-γ)/2}\]
\[Hence,\ TV^{(1-γ)/2} = constant\]