Question

(a)
When 117 J of energy is supplied as heat to 2.00 moles of an ideal gas at constant
pressure, the temperature rises by 2.00 K. Calculate the molar heat capacity at constant
pressure, CP,m and the molar heat capacity at constant volume, CV,m for the gas. By
considering the two heat capacities, determine whether the gas is monatomic or diatomic.

(b)

Explain what isothermal processes and adiabatic processes are, and the differences
between them, taking care to explain the role, or otherwise, of the environment to which the
system under study is exposed.

(c)

A sample of 8.02 × 10−1 moles of nitrogen gas (with γ = 1.40) occupies a volume of
2.00 × 10−2m3 at a pressure of 1.00 × 105 Pa and a temperature of 300K. The sample is
adiabatically compressed to exactly half its original volume. Nitrogen behaves as an ideal gas
under these conditions.

(i) What is the change in entropy of the gas?
(ii) Show from the adiabatic condition and the equation of state that TV (γ−1) remains
constant, and hence determine the final temperature of the gas.

(d) The gas sample is now returned to its initial state and then isothermally compressed to
half its original volume.
(i) Find the change in entropy of the gas.
(ii) What is the change in internal energy of the gas?
(iii) What is the amount of heat transferred from the gas to its environment?
(iv) Calculate the amount of work done in compressing the gas.

Answer 1

ONE QUESTION AT A TIME PLEASE, AS PER THE RULES

(a)

Q_p = nC_pT

C_p = Q_p / nT

C_p = 117 / 2 * 2

C_p = 29.25 J/K mole

and

C_p - C_v = R

C_v = C_p - R

C_v = 29.25 - 8.314

C_v = 20.936 J/k mole

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(b)

Isothermal simply means constant temperature. So any thermodynamic process that occurs at constant temperature is isothermal process.

An adiabatic process, on the other hand, is any process in which there was no heat exchanged with the object or system being described. Although the transmission of heat is isolated in adiabatic process, the temperature within the system temp is not constant.

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c)

(i)

For an adiabatic process, where only mechanical work is done on the system, the entropy does not change

s = 0

(ii)

The change of internal energy is given by

dU = dQ + dW

where

dQ = 0 for adiabatic processes

dW = - p dV

change of the internal energy an ideal gas in system is given by:

dU = nC_v dT

Combine the expressions to:

nC_vdT = - pdV

from ideal law

p = nRT / V

C_v dT = - R·T / V dV

integrating this, we get

TV^R/Cv= constant

C_p - C_v = R

= 1 + R / C_v

so,

TV^{( - 1}) = constant

(iii)

T₂ = T₁ (V₁/V₂)^{- 1}

T₂ = 395 K