0-056 kg of nitrogen is enclosed in a vessel at a temperature of 27-C- How much heat has to be transferred to the gas to double the most probable speed of its molecules- R - 2 cal-mol-k

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First Law of Thermodynamics

0.056 kg of n...

Question

$0.056 k g$ of nitrogen is enclosed in a vessel at a temperature of $27^{\circ} C$ . How much heat has to be transferred to the gas to double the most probable speed of its molecules? (R $= 2$ cal/mol.k)

4500 c a l

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9000 c a l

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18000 c a l

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5000 c a l

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Solution

The correct option is C $9000 c a l$
Since the gas is enclosed in a cylinder, volume is constant.
We have,
$(Δ Q)_{V} = n C_{V} Δ T$
Number of moles, $n = \frac{m a s s}{m o l e c u l a r w e i g h t} = \frac{0.056 \times 10^{3}}{28} .$
Since nitrogen is diatomic molecule, there are 3 translational + 2 rotational degree of freedom,
$C_{V} = (\frac{5}{2}) R$
We have $v \propto \sqrt{T}$ ( $v^{2} = \frac{3 k T}{m}$ )
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = 2$
$⟹ T_{2} = 4 T_{1}$
$Δ T = 4 T_{1} - T_{1} = 3 T_{1}$
$T_{1} = 27 + 273 = 300 K$
$⟹ Δ T = 3 \times 300 = 900 K$
Therefore,
$(Δ Q)_{V} = 2 \times \frac{5}{2} \times 2 \times 900$
$(Δ Q)_{V} = 9000 c a l$

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0.056kg of nitrogen is enclosed in a vessel at a temperature of 27∘C. How much heat has to be transferred to the gas to double the most probable speed of its molecules? (R =2 cal/mol.k)

$0.056 k g$ of nitrogen is enclosed in a vessel at a temperature of $27^{\circ} C$ . How much heat has to be transferred to the gas to double the most probable speed of its molecules? (R $= 2$ cal/mol.k)