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Heat and Thermodynamics question

2022 · 24 Jun · Shift 1 · Q62

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Heat and Thermodynamics
2022 · 24 Jun · Shift 1 · Q62

Heat and Thermodynamics question

2022 · 24 Jun · Shift 1 · Q62

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127

^\circ

C. Th amount of heat required to double the speed of its molecules is

\underline{\hspace{2cm}}

k cal. Take R = 2 cal mole

-

1 K

-

1)

Numerical answer

View written solutionFree

Correct answer: 12

Given data

Mass of nitrogen gas: $m = 0.056\,\text{kg} = 56\,\text{g}$
Initial temperature: $T_1 = 127^\circ\text{C} = 400\,\text{K}$
Gas: nitrogen $N_2$
Molar mass of $N_2 = 28\,\text{g/mol}$
Given: $R = 2\,\text{cal mol}^{-1}\text{K}^{-1}$

Find number of moles

n = \frac{56}{28} = 2\,\text{mol}

Relation between molecular speed and temperature

For an ideal gas, rms speed is proportional to $\sqrt{T}$ :

v \propto \sqrt{T}

If the speed is doubled,

\frac{v_2}{v_1} = 2 = \sqrt{\frac{T_2}{T_1}}

Squaring both sides,

\frac{T_2}{T_1} = 4

So,

T_2 = 4T_1 = 4 \times 400 = 1600\,\text{K}

Thus,

\Delta T = T_2 - T_1 = 1600 - 400 = 1200\,\text{K}

Heat required

Since the gas is enclosed in a vessel, volume is constant. Therefore,

Q = n C_V \Delta T

For diatomic gas $N_2$ ,

C_V = \frac{5}{2}R

Given $R = 2\,\text{cal mol}^{-1}\text{K}^{-1}$ ,

C_V = \frac{5}{2} \times 2 = 5\,\text{cal mol}^{-1}\text{K}^{-1}

Hence,

Q = 2 \times 5 \times 1200 = 12000\,\text{cal}

Convert to kcal:

Q = \frac{12000}{1000} = 12\,\text{kcal}

Final answer

\boxed{12}

Comparison with stored answer

Stored correct answer = $12$

Our derived answer also equals $12$ , so it agrees.

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