Two moles of a monatomic gas is taken through a cyclic process starting from A- as shown in figure- The volume ratio are VB-VA-2 and VD-VA-4- The temperature T A at A is 27- C- On the basis of this statement- answer the following questions-The temperature of the gas at point B isA- 27- CB- 235- CC- 327- CD- 300- C

The correct option is C 327^∘CProcess A B: V/T = constant or V_0/(273 + 27 )=2V_0/T_B or T_B = 600 K = 327^∘ C ...

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Standard XII

Physics

1st Law of Thermodynamics

Two moles of ...

Question

Two moles of a monatomic gas is taken through a cyclic process starting from A, as shown in figure. The volume ratio are $\frac{V_{B}}{V_{A}} = 2$ and $\frac{V_{D}}{V_{A}} = 4$ . The temperature $T_{A}$ at A is $27^{\circ}$ C. On the basis of this statement, answer the following questions.

The temperature of the gas at point B is

27^{\circ}

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235^{\circ}

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327^{\circ}

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300^{\circ}

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Solution

The correct option is C $327^{\circ}$ C
Process A - B: $\frac{V}{T}$ = constant

or $\frac{V_{0}}{(273 + 27)} = \frac{2 V_{0}}{T_{B}}$
$o r T_{B} = 600 K = 327^{\circ}$ C

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Two moles of a monatomic gas is taken through a cyclic process starting from A, as shown in figure. The volume ratio are VBVA=2 and VDVA=4. The temperature TA at A is 27∘ C. On the basis of this statement, answer the following questions. The temperature of the gas at point B is

The correct option is C 327∘CProcess A - B: VT = constant or V0(273+27)=2V0TB or TB=600K=327∘ C

The correct option is C $327^{\circ}$ C
Process A - B: $\frac{V}{T}$ = constant

or $\frac{V_{0}}{(273 + 27)} = \frac{2 V_{0}}{T_{B}}$
$o r T_{B} = 600 K = 327^{\circ}$ C