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1 mole of an ideal gas at 25°C is subjected to expansion reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in J k–1 mol–1)
  • a)
    19.15
  • b)
    – 19.15
  • c)
    4.7
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?
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1 mole of an ideal gas at 25°C is subjected to expansion reversibl...
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1 mole of an ideal gas at 25°C is subjected to expansion reversibl...
°C and 1 atm occupies 24.79 L.

This value is obtained using the ideal gas law, which relates the pressure, volume, temperature, and number of moles of an ideal gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. At 25°C, which is 298 K, and 1 atm, the ideal gas law becomes:

(1 atm) V = (1 mol) (0.08206 L·atm/mol·K) (298 K)

Solving for V gives:

V = 24.79 L

Therefore, 1 mole of an ideal gas at 25°C and 1 atm occupies 24.79 L.
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Community Answer
1 mole of an ideal gas at 25°C is subjected to expansion reversibl...
We know me that q = w = 2.303 nRt log v1/v2
∆S = q / t
for adiabetic process q = 0
so , ∆S = 0/ t =0
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1 mole of an ideal gas at 25°C is subjected to expansion reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in J k–1 mol–1)a)19.15b)– 19.15c)4.7d)zeroCorrect answer is option 'D'. Can you explain this answer?
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