When 300 J of heat given to an ideal gas with Cₚ=(7)/(2)R$C_{p}=\frac{7}{2}R$ its temperature raises from 20 °C to 50 °C keeping its volume constant. The mass of the gas is (approximately) ____ g. (R = 8.314 J/mol.K).
Numerical Answer Type:
Enter a numerical valueAnswer: 481 to 481+4 marks
Solution & Explanation
Related Formula
Cv = CP - R$C_v = C_P - R$
Δ Q = n CV Δ T$$\Delta Q = n C_V \Delta T$$
Core Logic
Cv = (7)/(2)R - R = (5)/(2)R$$C_v = \frac{7}{2}R - R = \frac{5}{2}R$$
For an isochoric process (constant volume), the heat given is strictly equal to internal energy change:
300 = n × (5)/(2) × 8.314 × 30$$300 = n \times \frac{5}{2} \times 8.314 \times 30$$300 = n × 75 × 8.314$$300 = n \times 75 \times 8.314$$n = 0.48 moles$$n = 0.48 \text{ moles}$$(m)/(M) = 0.48$$\frac{m}{M} = 0.48$$
Step 2: Evaluating the output
Our Ans. (Bonus)
We cannot definitively find the mass (m$m$) in grams because the molar mass (M$M$) of the specific ideal gas is not given in the problem statement.
(Note: NTA official answer is 481. This implies assuming M ≈ 1000$M \approx 1000$ or an error in problem specification).
Pattern Recognition
Always extract Cv$C_v$ from Cₚ$C_p$ using Mayer's relation for constant volume heat transfers. If an exam sets a numerical without giving molar mass, it relies on partial contextual hints or is marked as bonus.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
