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Q.A perfect gas at 27°C is heated at constant pressure then its volume becomes thrice times, the temperature of the gas will be

(a) 81°C
(b) 27°C
(c) 9°C
(d) 45°C
Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025MCQ· 1mImportance★★★★★
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At constant pressure, Charles's law says V∝TKelvinV \propto T_{Kelvin}. Tripling the volume rigorously requires TT (in kelvin) to triple: 300 K→900 K=627∘C300\text{ K} \to 900\text{ K} = 627^\circ\text{C} — a value not listed among the given options.

For an ideal ("perfect") gas at constant pressure, Charles's law applies:

V1T1=V2T2(T in kelvin, always)\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad (T \text{ in kelvin, always})

Initial temperature: T1=27∘C=300T_1 = 27^\circ\text{C} = 300 K. If the volume becomes three times as large, V2=3V1V_2 = 3V_1, then:

T2=3×T1=3×300 K=900 K=900−273=627∘CT_2 = 3 \times T_1 = 3 \times 300\text{ K} = 900\text{ K} = 900 - 273 = 627^\circ\text{C}

This is the physically correct result, but it does not match any of the four printed options (81, 27, 9, 45 °C). A very common slip when setting this kind of question is to multiply the Celsius reading directly by the volume ratio without first converting to kelvin — that is, 27∘C×3=81∘C27^\circ\text{C} \times 3 = 81^\circ\text{C}, which exactly matches option (a). This strongly suggests the paper's intended answer is (a), even though it is not the rigorous physics answer.

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