The molar specific heat of a diatomic ideal gas at constant volume is Cv = 5/2 R. What is the heat needed to heat 28.00 g of nitrogen gas from 20.00 °C to 85.00 °C?
Question
The molar specific heat of a diatomic ideal gas at constant volume is Cv = 5/2 R. What is the heat needed to heat 28.00 g of nitrogen gas from 20.00 °C to 85.00 °C?
Solution
To solve this problem, we need to use the formula for heat transfer:
q = nCvΔT
where:
- q is the heat transferred,
- n is the number of moles,
- Cv is the molar specific heat at constant volume, and
- ΔT is the change in temperature.
Step 1: Convert the mass of nitrogen gas to moles. Nitrogen (N2) has a molar mass of approximately 28.00 g/mol. So, 28.00 g of nitrogen gas is 1 mole.
Step 2: Convert the temperatures from Celsius to Kelvin. The Kelvin scale is used in thermodynamic calculations. The conversion is K = °C + 273.15. So, 20.00 °C = 293.15 K and 85.00 °C = 358.15 K.
Step 3: Calculate the change in temperature (ΔT). ΔT = T_final - T_initial = 358.15 K - 293.15 K = 65 K.
Step 4: Substitute the values into the heat transfer formula. The molar specific heat at constant volume (Cv) for a diatomic ideal gas is given as 5/2 R. The value of R (the gas constant) is 8.314 J/(mol*K).
q = nCvΔT = 1 mol * (5/2 * 8.314 J/(mol*K)) * 65 K = 1352.725 J
So, the heat needed to heat 28.00 g of nitrogen gas from 20.00 °C to 85.00 °C is approximately 1353 J.
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