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  • Three moles of an ideal gas are initially at a temperature T0 = 273 K. The gas is isothermally expanded to 5.0 times its initial volume (V1 → V2), and then isochorically heated so that the pre

Three moles of an ideal gas are initially at a temperature T0 = 273 K. The gas is isothermally expanded to 5.0 times its initial volume (V1 → V2), and then isochorically heated so that the pressure in the final state becomes equal to the initial state. The total amount of heat transferred to the gas during the process is Q = 80 kJ. Find the ratio P V γ = C/C for this gas, where γ is the specific heat ratio.

Option: 1

300.3


Option: 2

2541.36


Option: 3

7458.21


Option: 4

2156.25


Answers (1)

best_answer

Given that the gas undergoes two processes: an isothermal expansion and an isochoric heating. We are asked to find the ratio \frac{P V^\gamma}{C} for the gas. For an ideal gas, we have the equation of state: P V = nRT, where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Step 1: Isothermal Expansion During an isothermal process, the equation P V = nRT can be rewritten as:

P_1 V_1=P_2 V_2

where P1, V1 are the initial pressure and volume, and P2, V2 are the final pressure and volume. Given that V2 = 5V1, we have:

\begin{gathered} P_1 V_1=P_2 \cdot 5 V_1 \\ P_2=\frac{1}{5} P_1 \end{gathered}

Step 2: Isochoric Heating

In an isochoric process, the heat added is given by the equation:

Q=n C_v \Delta T

where Cv is the molar specific heat at constant volume and ?T is the change in temperature. Since the pressure in the final state becomes equal to the initial state, the temperature also becomes the same, i.e., T2 = T0. Thus, the heat added during isochoric heating is:

Q=n C_v\left(T_2-T_0\right)=n C_v\left(T_0-T_0\right)=0

This means that no heat is added during the isochoric process.

Step 3: Total Heat Transferred

The total heat transferred, Q, is the sum of the heat added during isothermal expansion and isochoric heating:

Q=Q_{\text {isothermal }}+Q_{\text {isochoric }}=Q_{\text {isothermal }}

\text { Therefore, } Q=n R T_0 \ln \left(\frac{V_2}{V_1}\right) \text {. }

Given that Q = 80 kJ and n = 3 moles, we can solve for R:

80 × 103 = 3R · 273 ln(5) Solving for R, we get: R ≈ 8.314 J/mol K

Step 4: Finding Specific Heat Ratio The specific heat ratio, γ, for an ideal gas is given by:

\gamma=\frac{C_p}{C_v}

where Cp is the molar specific heat at constant pressure, and Cv is the molar specific heat at constant volume.

\text { For an ideal diatomic gas, } \gamma=\frac{7}{5}=1.4

Step 5: Finding the Ratio \frac{{P} V^\gamma}{C}

\text { The ratio } \frac{P V^\gamma}{C} \text { can be expressed in terms of known quantities: }\begin{aligned} &\frac{P V^\gamma}{C}=\frac{P_1 V_1^\gamma}{n R}=\frac{P_1 V_1^\gamma}{3 R}\\ &P V=n R T . \end{aligned}

Using the ideal gas law P V = nRT:

\frac{P_1 V_1^\gamma}{3 R}=\frac{n R T_0 V_1^\gamma}{3 R V_1}=\frac{T_0 n V_1^{\gamma-1}}{3}

Substituting γ = 1.4 and V2 = 5V1:

\frac{T_0 n V_1^{0.4}}{3}=\frac{273 \times 3 \times(1)^{0.4}}{3}=273 \times 1.1 \approx 300.3Therefore, the ratio \frac{P V^\gamma}{C} is approximately 300.3. Therefore, the correct option is 1

 

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Rakesh

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