# Isochoric thermodynamics processes – problems and solutions

1. PV diagram below shows an ideal gas

Solution :

Process AB is an isochoric process (constant volume). The volume is constant so that no work is done by the gas.

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2. Three moles of monoatomic gas at 47^{o}C and at pressure 2 x 10^{5} Pa, undergoes isochoric process so that pressure increases 3 x 10^{5} Pa. The change in internal energy of the gas is… Universal gas constant (R) = 8.315 J/mol.K

__Known :__

Initial temperature (T_{1}) = 47^{o}C + 273 = 320 K

Initial pressure (P_{1}) = 2 x 10^{5} Pa

Final pressure (P_{2}) = 3 x 10^{5} Pa

Universal gas constant (R) = 8.315 J/mol.K

Number of moles (n) = 3

__Wanted:__ The change in internal energy of the gas.

__Solution :__

In the isochoric process, the volume is kept constant so that no work is done by the gas (W = 0).

**The first law of thermodynamics :**

ΔU = Q-W

ΔU = Q-0

ΔU = Q

*ΔU = internal energy, Q = heat*

__Internal energy of gas :__

ΔU = 3/2 n R ΔT = 3/2 n R (T_{2} – T_{1})

__Gay-Lussac____‘s law____ (____constant volume____)__ :

The change in internal energy of gas :

ΔU = 3/2 n R (T_{2} – T_{1}) = 3/2 (3)(8.315)(480-320)

ΔU = 3/2 (24.945)(160) = 3/2 (3991.2)

ΔU = 5986.8 Joule

3. 0.2 moles of monatomic gases at 27^{o}C are in a closed container. The heat

__Known :__

Number of moles (n) = 0.2 mol

Initial temperature (T_{1}) = 27^{o}C + 273 = 300 K

Final temperature (T_{2}) = 400 K

Universal constant gas (R) = 8.315 J/mol.K

__Wanted ____:__ Heat is added (Q)

__Solution :__

In isochoric process, volume is kept constant so that no work is done by the gas (W = 0).

**The first law of thermodynamics :**

ΔU = Q-W

ΔU = Q-0

ΔU = Q

*ΔU = internal energy, Q = heat*

The internal energy of gas :

ΔU = 3/2 n R ΔT = 3/2 n R (T_{2} – T_{1})

ΔU = 3/2 (0.2)(8.315)(400-300)

ΔU = 3/2 (0.2)(8.315)(100)

ΔU = 249.45 Joule