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4. 48 L of an ideal gas at stp requires 12 cal to raise its temperature by 15 degree c at constant volume. the Cp of the gas is.?
Most Upvoted Answer
4. 48 L of an ideal gas at stp requires 12 cal to raise its temperatur...
No. of moles in 4.484 of ideal gas at STP = 4.4822.4=0.24.4822.4=0.2
Thus to raise the temperature of 0.2 mol of the ideal gas, through 15∘C15∘C heat absorbed = 12 cal.
∴∴ to raise the temperature of 1 mol of the gas through 1∘C1∘C, heat absorbed = 1215×10.2=41215×10.2=4 cal
i.e., CVCV = 4 cal
∴CP=CV+R∴CP=CV+R = 4 cal + 2 cal = 6 cal
Community Answer
4. 48 L of an ideal gas at stp requires 12 cal to raise its temperatur...
Understanding the Problem
To find the specific heat at constant pressure (Cp) of the gas, we start with the given information:
- Volume of the gas = 48 L
- Heat added (Q) = 12 cal
- Temperature change (ΔT) = 15 °C
- Conditions = Standard Temperature and Pressure (STP)
Finding the Moles of the Gas
At STP, 1 mole of an ideal gas occupies 22.4 L. Thus, the number of moles (n) of the gas can be calculated as follows:
- n = Volume / Molar volume at STP
- n = 48 L / 22.4 L/mol = 2.14 moles (approximately)
Calculating the Heat Capacity at Constant Volume (Cv)
The heat added at constant volume is given by the equation:
- Q = n * Cv * ΔT
Rearranging this gives us:
- Cv = Q / (n * ΔT)
Substituting the known values:
- Cv = 12 cal / (2.14 mol * 15 °C)
- Cv ≈ 0.38 cal/(mol·°C)
Relating Cp and Cv
For ideal gases, the relationship between Cp and Cv is given by:
- Cp = Cv + R
Where R is the universal gas constant. In calories, R ≈ 2 cal/(mol·°C).
Final Calculation of Cp
Using the value of Cv we found:
- Cp = 0.38 cal/(mol·°C) + 2 cal/(mol·°C)
- Cp ≈ 2.38 cal/(mol·°C)
Conclusion
Thus, the specific heat at constant pressure (Cp) of the gas is approximately:
- Cp ≈ 2.38 cal/(mol·°C)
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4. 48 L of an ideal gas at stp requires 12 cal to raise its temperature by 15 degree c at constant volume. the Cp of the gas is.?
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