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2 mole of an ideal gas at 27ºC temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) [2002]
- a)92.1
- b)0
- c)4
- d)9.2
Correct answer is option 'D'. Can you explain this answer?
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2 mole of an ideal gas at 27ºC temperature is expanded reversibly...
For isothermal reversible expansion
Entropy change, 
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2 mole of an ideal gas at 27ºC temperature is expanded reversibly...
To calculate the volume of 2 moles of an ideal gas at 27°C, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in K)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 27 + 273.15
T(K) = 300.15 K
Next, we can plug in the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (2 mol * 0.0821 L·atm/mol·K * 300.15 K) / P
The pressure (P) is not given in the question, so we cannot calculate the exact volume without that information.
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in K)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 27 + 273.15
T(K) = 300.15 K
Next, we can plug in the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (2 mol * 0.0821 L·atm/mol·K * 300.15 K) / P
The pressure (P) is not given in the question, so we cannot calculate the exact volume without that information.
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2 mole of an ideal gas at 27ºC temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) [2002]a)92.1b)0c)4d)9.2Correct answer is option 'D'. Can you explain this answer?
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