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Heat and Thermodynamics question

2022 · 29 Jun · Shift 1 · Q45
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  5. /2022 · 29 Jun · Shift 1 · Q45

Heat and Thermodynamics question

2022 · 29 Jun · Shift 1 · Q45

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0 ∘^\circ∘ C will be : (Given gas constant R = 8.3 JK −-− 1-mol −-− 1)
  1. A
    249 J
  2. B
    415 J
  3. C
    498 J
  4. D
    830 J
View written solutionFree

Correct answer: C

  1. Identify the process

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

Hence, heat supplied is Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT

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

  1. Find the number of moles of helium

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

Given volume: V=44.8 LV = 44.8\text{ L}V=44.8 L

So, n=44.822.4=2 moln = \frac{44.8}{22.4} = 2 \text{ mol}n=22.444.8​=2 mol

  1. Temperature rise

Given increase in temperature: ΔT=20∘C=20 K\Delta T = 20^\circ \text{C} = 20\text{ K}ΔT=20∘C=20 K

  1. Compute the heat required

Using Q=nCVΔT=n(32R)ΔTQ = nC_V\Delta T = n\left(\frac{3}{2}R\right)\Delta TQ=nCV​ΔT=n(23​R)ΔT

Substitute values: Q=2×32×8.3×20Q = 2 \times \frac{3}{2} \times 8.3 \times 20Q=2×23​×8.3×20

Q=3×8.3×20Q = 3 \times 8.3 \times 20Q=3×8.3×20

Q=24.9×20=498 JQ = 24.9 \times 20 = 498\text{ J}Q=24.9×20=498 J

  1. Match with the options

Q=498 JQ = 498\text{ J}Q=498 J

So the correct option is C.

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