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When an ideal diatomic gas is heated at ...

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied, which increases the internal energy of the gas, is

A

`(2)/(5)`

B

`(3)/(5)`

C

`(3)/(7)`

D

`(5)/(7)`

Text Solution

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The correct Answer is:
To solve the question regarding the fraction of heat energy supplied that increases the internal energy of an ideal diatomic gas when heated at constant pressure, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Process**: - The process described is isobaric (constant pressure). For an ideal gas, the relationship between heat supplied (Q), change in internal energy (ΔU), and work done (W) can be expressed using the first law of thermodynamics: \[ Q = ΔU + W \] 2. **Identify the Work Done**: - For an isobaric process, the work done (W) is given by: \[ W = P \Delta V \] - Using the ideal gas law, we can express this in terms of temperature change (ΔT): \[ W = nR \Delta T \] - Here, \(n\) is the number of moles, \(R\) is the gas constant, and \(\Delta T\) is the change in temperature. 3. **Change in Internal Energy**: - The change in internal energy (ΔU) for a diatomic ideal gas can be expressed as: \[ ΔU = \frac{f}{2} n R \Delta T \] - For a diatomic gas, the degrees of freedom \(f = 5\) (3 translational and 2 rotational). 4. **Substituting Values**: - Substitute \(f = 5\) into the equation for ΔU: \[ ΔU = \frac{5}{2} n R \Delta T \] 5. **Relating Q, ΔU, and W**: - Now we can express Q using the equations for ΔU and W: \[ Q = ΔU + W = \frac{5}{2} n R \Delta T + n R \Delta T \] - Simplifying this gives: \[ Q = \left(\frac{5}{2} + 1\right) n R \Delta T = \frac{7}{2} n R \Delta T \] 6. **Finding the Fraction**: - The fraction of heat supplied that increases the internal energy is given by: \[ \text{Fraction} = \frac{ΔU}{Q} = \frac{\frac{5}{2} n R \Delta T}{\frac{7}{2} n R \Delta T} \] - Simplifying this fraction: \[ \text{Fraction} = \frac{5/2}{7/2} = \frac{5}{7} \] ### Final Answer: The fraction of the heat energy supplied that increases the internal energy of the gas is \(\frac{5}{7}\).
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