Isothermal and Adiabatic Processes Numerical on Molar Specific Heat Class-11 Nootan ISC Physics 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 on Molar Specific Heat Class-11
(Nootan ISC Physics Solutions Ch-20)
Board | ISC |
Class | 11 |
Subject | Physics |
Book | Nootan |
Chapter-20 | Isothermal and Adiabatic Processes |
Topics | Numerical on Molar Specific Heat |
Academic Session | 2024-2025 |
Isothermal and Adiabatic Processes Numerical on Molar Specific Heat
page-780
Que-14: 0.5 mole CO2 is heated from 25°C to 125°C at constant pressure. If Cp = 8.0 Cal / mol-°C and Cv = 6.0 Cal / mol-°C for CO2, then calculate (i) amount of heat given to the gas, (ii) increase in the internal energy of the gas.
Ans: Cp – Cv = R
=> μ Cp ΔT – μ Cv ΔT = R ΔT
=> ΔQ – ΔU = R ΔT
=> ΔQ = 0.5 x 8 x 100 = 400 Cal
ΔU = μ Cv ΔT = 0.5 (Cp – R) ΔT
=> 0.5 x (8 – 2) x 100 = 300 Cal
Que-15: 0.1 mole nitrogen is heated from 27°C to 327°C at constant pressure. Determine the heat given to the gas, increase in internal energy of the gas and work done by the gas. For nitrogen Cp = 7 and C, = 5 Cal / mol-°C
Ans: ΔQ = μ Cp ΔT
=> 0.1 x 7 x 300 = 210 Cal
ΔU = μ Cv ΔT
=> 0.1 (7 – 2) x 300 = 150 Cal
ΔW = μ R ΔT
=> 0.1 x 2 x 300 = 60 Cal
Que-16: Calculate the amount of heat required to raise the temperature of two moles of an ideal monoatomic gas from 0°C to 100°C if no work is done. You are given that the molar specific heat of the above gas at constant pressure is 2.5 R, where R is the universal gas constant (R = 8.3 J / mol-K).
Ans: ΔQ = μ Cv ΔT
=> 2 x (Cp – R) ΔT
=> 2 x (2.5 R – R) x ΔT
=> 2 x 1.5 R x ΔT
=> 2 x 1.5 x 8.31 x 100
=> 2490 J
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