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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Specific Heat Capacities of Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Specific Heat Capacities of Gases

An ideal gas ( γ = 1.67 ) is taken through the process abc shown in...

An ideal gas (γ=1.67\gamma = 1.67) is taken through the process abcabc shown in the figure. The temperature at the point aa is 300 K300 \text{ K}. Calculate the temperatures at bb and cc.
An ideal gas ( γ = 1.67 ) is taken through the process abc shown in the figure. The temperature at the point a is 300 K

Options

  1. A600 K600 \text{ K}, 900 K900 \text{ K}
  2. B600 K600 \text{ K}, 1200 K1200 \text{ K}
  3. C900 K900 \text{ K}, 600 K600 \text{ K}
  4. D300 K300 \text{ K}, 450 K450 \text{ K}

Correct answer

A. 600 K600 \text{ K}, 900 K900 \text{ K}

Step-by-step solution

From the given graph, the pressure and volume at points aa, bb, and cc are:
Pa=100 kPaP_a = 100 \text{ kPa}, Va=100 cm3V_a = 100 \text{ cm}^3
Pb=200 kPaP_b = 200 \text{ kPa}, Vb=100 cm3V_b = 100 \text{ cm}^3
Pc=200 kPaP_c = 200 \text{ kPa}, Vc=150 cm3V_c = 150 \text{ cm}^3

Using the ideal gas equation, PVT=constant\dfrac{PV}{T} = \text{constant}.

For the isochoric process a→ba \to b (V=constantV = \text{constant}):
PaTa=PbTb\dfrac{P_a}{T_a} = \dfrac{P_b}{T_b}
⇒Tb=Ta×PbPa=300×200100=600 K\Rightarrow T_b = T_a \times \dfrac{P_b}{P_a} = 300 \times \dfrac{200}{100} = 600 \text{ K}

For the isobaric process b→cb \to c (P=constantP = \text{constant}):
VbTb=VcTc\dfrac{V_b}{T_b} = \dfrac{V_c}{T_c}
⇒Tc=Tb×VcVb=600×150100=900 K\Rightarrow T_c = T_b \times \dfrac{V_c}{V_b} = 600 \times \dfrac{150}{100} = 900 \text{ K}

The temperatures at bb and cc are 600 K600 \text{ K} and 900 K900 \text{ K} respectively.
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