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Heat and Thermodynamics question

2019 · 10 Apr · Shift 1 · Q57
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Heat and Thermodynamics question

2019 · 10 Apr · Shift 1 · Q57

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is : [Given that R = 8.31 J mol–1 K–1]
  1. A
    374 J
  2. B
    700 J
  3. C
    748 J
  4. D
    350 J
View written solutionFree

Correct answer: C

  1. Identify the process

    The cylinder has fixed capacity, so the gas is heated at constant volume.

    Therefore, Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT

  2. Find number of moles of helium

    At STP, 111 mole of an ideal gas occupies 22.422.422.4 L.

    Given volume, V=67.2 LV = 67.2\text{ L}V=67.2 L

    So number of moles is n=67.222.4=3n = \frac{67.2}{22.4} = 3n=22.467.2​=3

  3. Heat capacity of helium at constant volume

    Helium is a monoatomic ideal gas, so CV=32RC_V = \frac{3}{2}RCV​=23​R

    Given, R=8.31 J mol−1K−1R = 8.31\, \text{J mol}^{-1}\text{K}^{-1}R=8.31J mol−1K−1

    Hence, CV=32(8.31)=12.465 J mol−1K−1C_V = \frac{3}{2}(8.31) = 12.465\, \text{J mol}^{-1}\text{K}^{-1}CV​=23​(8.31)=12.465J mol−1K−1

  4. Temperature rise

    ΔT=20∘C=20 K\Delta T = 20^\circ \text{C} = 20\text{ K}ΔT=20∘C=20 K

  5. Calculate heat supplied

    Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT Q=3×12.465×20Q = 3 \times 12.465 \times 20Q=3×12.465×20 Q=747.9 JQ = 747.9\, \text{J}Q=747.9J

    Q≈748 JQ \approx 748\, \text{J}Q≈748J

  6. Check options

    • A: 374374374 J ❌
    • B: 700700700 J ❌
    • C: 748748748 J ✅
    • D: 350350350 J ❌

Thus, the correct answer is Option C.

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