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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

A gas is confined in a cylindrical can fitted with a piston. Both...

A gas is confined in a cylindrical can fitted with a piston. Both the piston and the walls of the can are adiabatic. The initial pressure, volume, and temperature of the gas are 100 kPa100 \text{ kPa}, 400 cm3400 \text{ cm}^3, and 300 K300 \text{ K} respectively. The ratio of the specific heat capacities of the gas is Cp/CV=1.5C_p / C_V = 1.5. Calculate the pressure and the temperature of the gas if it is suddenly compressed to a volume of 100 cm3100 \text{ cm}^3.

Options

  1. A800 kPa800 \text{ kPa} and 2400 K2400 \text{ K}
  2. B800 kPa800 \text{ kPa} and 600 K600 \text{ K}
  3. C400 kPa400 \text{ kPa} and 300 K300 \text{ K}
  4. D400 kPa400 \text{ kPa} and 600 K600 \text{ K}

Correct answer

B. 800 kPa800 \text{ kPa} and 600 K600 \text{ K}

Step-by-step solution

For an adiabatic process, the relationship between pressure and volume is given by P1V1γ=P2V2γP_1 V_1^{\gamma} = P_2 V_2^{\gamma}.

Substituting the given values:

P2=P1(V1V2)γP_2 = P_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma}

P2=100×(400100)1.5P_2 = 100 \times \left( \dfrac{400}{100} \right)^{1.5}

P2=100×41.5=100×8=800 kPaP_2 = 100 \times 4^{1.5} = 100 \times 8 = 800 \text{ kPa}

The relationship between temperature and volume for an adiabatic process is T1V1γ−1=T2V2γ−1T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}.

Substituting the given values:

T2=T1(V1V2)γ−1T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}

T2=300×(400100)1.5−1T_2 = 300 \times \left( \dfrac{400}{100} \right)^{1.5 - 1}

T2=300×40.5=300×2=600 KT_2 = 300 \times 4^{0.5} = 300 \times 2 = 600 \text{ K}
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