Isothermal and Adiabatic Processes Numerical Class-11Physics Nootan Solutions Ch-20. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Isothermal and Adiabatic Processes Numerical Class-11Physics Nootan Solutions
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-20 | Isothermal and Adiabatic Processes |
| Topics | Numerical on Isothermal and Adiabatic Processes |
| Academic Session | 2024-2025 |
Numerical on Isothermal and Adiabatic Processes
Que-1: The volume of an ideal gas in a vessel is 2 liters at normal pressure (1.0 × 10^5 N/m²). The pressure of the gas is increased (i) under isothermal condition, (ii) under adiabatic condition until its volume remains 1 liter. Find the increased pressure of the gas.
Ans: (i) For isothermal process
=> P1 V1 = P2 V2
=> 1 x 10^5 x 2 = P2 x 1
=> P2 = 2.0 x 10^5 N/m²
(ii) For adiabatic process
=> P1 V1^γ = P2 V2^γ
=> (V1 / V2)^γ = P2 / P1 = (P2 / 1 x 10^5) = (2/1)^(1.4)
=> P2 = 2.64 x 10^5 N/m²
Que-2: The initial pressure of a gas is 5 x 10^5 N/m². Its volume is compressed to 1/9 of its original volume adiabatically. What will be the pressure of the gas in this condition? For the gas γ = 3/2.
Ans: For adiabatic process
=> P1 V1^γ = P2 V2^γ
=> P2 / P1 = (V1 / V2)^(3/2) = [V1 / (V1/9)]^(3/2) = 27
=> P2 = P1 x 27
=> 5 x 10^5 x 27
=> 1.35 x 10^7 n/m²
Que-3: A vessel is filled with a perfect gas at a pressure of 2.0 x 10^5 N/m². The pressure of the gas is increased adiabatically to such an extent that its volume is reduced to half of its original volume. Calculate the new pressure. Given γ = 1.41 and (2)^1.41 = 2.66.
Ans: For adiabatic process
=> P1 V1^γ = P2 V2^γ
=> P2 / P1 = (V1 / V2)^γ
=> P2 / (2 x 10^5) = [V1 / (V1/2)]^(1.41) = (2)^1.41 = 2.66
=> P2 = 5.32 x 10^5 N/m²
Que-4: The volume of a given mass of air is doubled in adiabatic expansion. If the initial pressure of the air is 80 cm of mercury, what will be the final pressure ? γ for air = 1.41 and (2)^1.41 = 2.66
Ans: For adiabatic process
=> P2 / P1 = (V1 / V2)^γ
=> P2 / 80 = (V1 / 2V1)^(1.41) = (1/2)^(1.41)
=> P2 = 80 / 2.66 = 30 cm of hg
Que-5: One liter of a gas whose initial pressure is 1 atmosphere is compressed till the pressure becomes 2 atmospheres. If the gas be compressed (i) slowly, (ii) suddenly, what will be the new volumes of the gas? γ = 1.4 and (0.5)^(1/1.4) = 0.61.
Ans: (i) Slowly means isothermal process
=> P1 V1 = P2 V2
=> 1 x 1 = 2 x V2
=> V2 = 0.50 liter
(ii) Sudden process is adiabatic process
=> P1 / P2 = (V2 / V1)^γ
=> (V2 / V1) = (P1 / P2)^(1/γ)
=>(V2/1) = (1/2)^(1/1.4)
=>(V2/1) = 0.61
=> V2 = 0.61 liter
Que-6: The pressure of a gas (γ = 1.5) is suddenly raised to 8 times. Calculate, how many times the volume of gas will become.
Ans: Sudden process shows adiabatic process
∴ (V1 / V2) = (P2 / P1)^(1/γ)
=> (V1 / V2) = 8^(2/3)
=> V2 = V1 / 4
V2 will become one-forth times of V1.
Que-7: A certain mass of air (γ = 1.5) at 27°C is compressed (i) slowly and (ii) suddenly to one-fourth of original volume. Find the final temperature of the compressed air in each case.
Ans: (i) a slow process is iso-thermal process in which temperature of system remains constant i.e. 27°
(ii) A sudden process is adiabatic process for which
=> T1 V1^(γ-1) = T2 V2^(γ-1)
=> (V1/V2)^(γ-1) = (T2/T1)
=> [V1/(V1/4)]^(1.5-1) = T2/ (273 + 27)
=> 2 = T2/300
=> T2 = 600 K
or 600 – 273 = 327 °C
Que-8: A gas initially at atmospheric pressure and 15°C temperature is compressed adiabatically until its volume remains one-fourth of its original volume. Find out the final pressure and temperature of the gas (γ = 1.5).
Ans: For adiabatic process
=> P1 V1^γ = P2 V2^γ
=> P1 / P2 = (V2 / V1)^γ
=> 1/P2 = (V1/4V1)^(3/2)
=> P2 = 8 atm
again, (V1/V2)^(γ-1) = (T2/T1)
=> [V1/(V1/4)]^(1.5-1) = T2/ (273 + 15)
=> 2 = T2/288
=> T2 = 576 K
or 576 – 273 = 303 °C
Que-9: A monoatomic ideal gas at 17°C is suddenly compressed to 1/8 of its initial volume. Find the final temperature of the gas. Given : for the monoatomic gas γ = 5/3.
Ans: For adiabatic process
=> (V1/V2)^(γ-1) = (T2/T1)
=> [V1/(V1/8)]^(5/3) = T2/(273 + 17)
=> T2 = 887 °C
Que-10: A gas, whose initial temperature is 27°C, is compressed adiabatically to 8 times its initial pressure. If γ = 1.5, find the rise in temperature in this transformation.
Ans: For adiabatic process
=> T1^γ / P1^(γ-1) = T2^γ / P2^(γ-1)
=> (T1 / T2)^γ = (P1 / P2)^(γ-1)
=> (2723 + 27/ T2) = (P1 / 8P1)^(1.5-1 / 1.5)
=> 300 / T2 = 1/2
=> T2 = 600 K
or 600 – 273 = 327 °C
— : end of Isothermal and Adiabatic Processes Numerical Class-11 Physics Nootan Solutions :–
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