Two moles of an ideal monoatomic gas occupies a volume $V$ at $27^{\circ}C$. The gas expands adiabatically to a volume $2V$. Calculate $(a)$ the final temperature of the gas and $(b) $ change in its internal energy -

Question

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion (A): B_eCl₂ and MgCl₂ produce characteristic flame.
Reason (R) : The excitation energy is high in B_eCl₂ and MgCl₂
In the light of the above statements, choose the correct answer from the option given below

Answer 1

To solve the given problem, we need to determine the final temperature and the change in internal energy of an ideal monoatomic gas undergoing adiabatic expansion.

Concepts and Formulae

An adiabatic process is one where no heat is exchanged with the surroundings, meaning $Q = 0$.
The relation between temperature and volume for an adiabatic process is given by: \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] where $\gamma$ (gamma) is the heat capacity ratio, $\frac{C_p}{C_v}$. For a monoatomic ideal gas, $\gamma = \frac{5}{3}$.
Change in internal energy ($ \Delta U $) for an ideal gas is given by: \[ \Delta U = n C_v \Delta T \] where $n$ is the number of moles and $C_v = \frac{3R}{2}$ for a monoatomic gas.

Step-by-Step Solution

Initial Conditions:
- Moles ($n$) = 2
- Initial Temperature ($T_1$) = $27^\circ C = 300 \, K$
- Initial Volume ($V_1$) = $V$
- Final Volume ($V_2$) = $2V$
- $\gamma = \frac{5}{3}$
Calculate final temperature ($T_2$) using the adiabatic process relation: \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] Substituting the known values: \[ 300 \times V^{\frac{2}{3}} = T_2 \times (2V)^{\frac{2}{3}} \] Simplify and solve for $T_2$: \[ 300 = T_2 \times 2^{\frac{2}{3}} \] \[ T_2 = \frac{300}{2^{\frac{2}{3}}} \] Calculation: \[ T_2 \approx 189.2 \text{ K} \] Rounding off, we get $T_2 = 189 \text{ K}$.
Calculate the change in internal energy ($\Delta U$): Using the formula: \[ \Delta U = n C_v \Delta T \] \[ C_v = \frac{3R}{2}, \quad \Delta T = T_2 - T_1 = 189 - 300 = -111 \text{ K} \] \[ \Delta U = 2 \times \frac{3R}{2} \times (-111) \] Simplifying: \[ \Delta U = -3R \times 111 \] Assuming $R = 8.314 \, \text{J/mol K}$: \[ \Delta U = -3 \times 8.314 \times 111 \approx -2.7 \text{ kJ} \]

Conclusion

The final temperature of the gas after adiabatic expansion is 189 K. The change in internal energy is -2.7 kJ. Thus, the correct answer is:

(a) 189 K (b) -2.7 kJ

This explanation aligns with the provided correct option.

Answer 2

To solve the given problem, we need to determine the final temperature and the change in internal energy of an ideal monoatomic gas undergoing adiabatic expansion.

Concepts and Formulae

An adiabatic process is one where no heat is exchanged with the surroundings, meaning $Q = 0$.
The relation between temperature and volume for an adiabatic process is given by: \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] where $\gamma$ (gamma) is the heat capacity ratio, $\frac{C_p}{C_v}$. For a monoatomic ideal gas, $\gamma = \frac{5}{3}$.
Change in internal energy ($ \Delta U $) for an ideal gas is given by: \[ \Delta U = n C_v \Delta T \] where $n$ is the number of moles and $C_v = \frac{3R}{2}$ for a monoatomic gas.

Step-by-Step Solution

Initial Conditions:
- Moles ($n$) = 2
- Initial Temperature ($T_1$) = $27^\circ C = 300 \, K$
- Initial Volume ($V_1$) = $V$
- Final Volume ($V_2$) = $2V$
- $\gamma = \frac{5}{3}$
Calculate final temperature ($T_2$) using the adiabatic process relation: \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] Substituting the known values: \[ 300 \times V^{\frac{2}{3}} = T_2 \times (2V)^{\frac{2}{3}} \] Simplify and solve for $T_2$: \[ 300 = T_2 \times 2^{\frac{2}{3}} \] \[ T_2 = \frac{300}{2^{\frac{2}{3}}} \] Calculation: \[ T_2 \approx 189.2 \text{ K} \] Rounding off, we get $T_2 = 189 \text{ K}$.
Calculate the change in internal energy ($\Delta U$): Using the formula: \[ \Delta U = n C_v \Delta T \] \[ C_v = \frac{3R}{2}, \quad \Delta T = T_2 - T_1 = 189 - 300 = -111 \text{ K} \] \[ \Delta U = 2 \times \frac{3R}{2} \times (-111) \] Simplifying: \[ \Delta U = -3R \times 111 \] Assuming $R = 8.314 \, \text{J/mol K}$: \[ \Delta U = -3 \times 8.314 \times 111 \approx -2.7 \text{ kJ} \]

Conclusion

The final temperature of the gas after adiabatic expansion is 189 K. The change in internal energy is -2.7 kJ. Thus, the correct answer is:

(a) 189 K (b) -2.7 kJ

This explanation aligns with the provided correct option.

Two moles of an ideal monoatomic gas occupies a volume $V$ at $27^{\circ}C$. The gas expands adiabatically to a volume $2V$. Calculate $(a)$ the final temperature of the gas and $(b) $ change in its internal energy -

The Correct Option is C

Solution and Explanation

Concepts and Formulae

Step-by-Step Solution

Conclusion

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