The H calories of heat is required to increase the temperature of one mole of monoatomic gas from 20∘C to 30∘C at constant volume. The quantity of heat required to increase the temperature of 2 moles of a diatomic gas from 20∘C to 25∘C is at constant volume is

For mono atomic gas CV=3R2Cv=1ndQdT⇒dQ=nCvdTH=1×3R2×10⇒H=15RFor diatomic gas CV=5R2Cv=1ndQdT⇒dQ=nCvdTdQ2=2×5R2×5⇒H=25RdQ2H=25R15R⇒dQ2=5H3

The H calories of heat is required to increase the temperature of one mole of monoatomic gas from 20∘C to 30∘C at constant volume. The quantity of heat required to increase the temperature of 2 moles of a diatomic gas from 20∘C to 25∘C is at constant volume is

For mono atomic gas CV=3R2Cv=1ndQdT⇒dQ=nCvdTH=1×3R2×10⇒H=15RFor diatomic gas CV=5R2Cv=1ndQdT⇒dQ=nCvdTdQ2=2×5R2×5⇒H=25RdQ2H=25R15R⇒dQ2=5H3

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The H calories of heat is required to increase the temperature of one mole of monoatomic gas from $20^{\circ} C$ to $30^{\circ} C$ at constant volume. The quantity of heat required to increase the temperature of 2 moles of a diatomic gas from $20^{\circ} C$ to $25^{\circ} C$ is at constant volume is

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$\frac{5 H}{3}$

$\frac{7 H}{3}$

$\frac{4 H}{3}$

$2 H$

answer is B.

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Detailed Solution

For mono atomic gas $C_{V} = \frac{3 R}{2}$

$\begin{array}{l} C_{v} = \frac{1}{n} \frac{d Q}{d T} \Rightarrow d Q = n C_{v} d T \\ H = 1 \times \frac{3 R}{2} \times 10 \Rightarrow H = 15 R \end{array}$

For diatomic gas $C_{V} = \frac{5 R}{2}$

$\begin{array}{l} C_{v} = \frac{1}{n} \frac{d Q}{d T} \Rightarrow d Q = n C_{v} d T \\ d Q_{2} = 2 \times \frac{5 R}{2} \times 5 \Rightarrow H = 25 R \end{array}$