One mole of an ideal monoatomic gas requires 207 J heat to raise the t...
We know that Qv = nCvΔT and Qp = nCpΔT
Given Qp = 207 J

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One mole of an ideal monoatomic gas requires 207 J heat to raise the t...
Given:
- Heat required to raise the temperature of 1 mole of an ideal monoatomic gas by 10 K at constant pressure = 207 J
- Gas constant, R = 8.3 J/mol.K
To find:
- Heat required to raise the temperature of the same gas by 10 K at constant volume
Solution:
1. Heat capacity at constant pressure (Cp) can be calculated as:
Cp = (ΔH / ΔT)p
where, ΔH = change in enthalpy, ΔT = change in temperature, and p = constant pressure
2. Heat capacity at constant volume (Cv) can be calculated as:
Cv = (ΔU / ΔT)v
where, ΔU = change in internal energy, and v = constant volume
For an ideal gas, internal energy depends only on temperature, so
ΔU = nCvΔT
where, n = number of moles of gas
3. The relationship between Cp and Cv for an ideal gas is:
Cp - Cv = nR
where, R = gas constant
Using these equations, we can find the heat required to raise the temperature of the gas at constant volume:
ΔU = nCvΔT
ΔH = ΔU + pΔV
where, ΔV = change in volume
At constant volume, ΔV = 0, so ΔH = ΔU
Cv = (ΔU / ΔT)v
ΔU = nCvΔT
Substituting these values,
Cv = (ΔU / ΔT)v = (ΔH / ΔT)v - pΔV / ΔT
Cp - Cv = nR
Cp = Cv + nR
ΔH = ΔU = nCvΔT
ΔH = nCpΔT - nRΔT
207 J = nCp(10 K) - nR(10 K)
nCp - nR = 20.7 J/mol
Cv = Cp - nR
Cv = 20.7 J/mol / n
For 1 mole of gas,
Cv = 20.7 J/mol
ΔU = nCvΔT = (1 mol)(20.7 J/mol)(10 K) = 207 J
Therefore, the heat required to raise the temperature of the gas by 10 K at constant volume is 207 J.
Answer: Option D (124 J) is incorrect.