At 298k the coefficient of thermal expansion of water is α=2.07 *10-⁴ k-¹ and the density is 0.9970g/m³.if the temperature of 200g of water is raised from 298k to 323k under a constant pressure of 101 kpa . calculate w and Q and ΔU given CP= 75.30j/K .mol?

Question

Answer

Calculation of work, heat and change in internal energy of water

Given

Coefficient of thermal expansion of water (α) = 2.07 * 10^-4 K^-1

Density of water (ρ) = 0.9970 g/m³

Initial temperature (T1) = 298 K

Final temperature (T2) = 323 K

Constant pressure (P) = 101 kPa

Mass of water (m) = 200 g

Specific heat capacity of water at constant pressure (CP) = 75.30 J/K.mol

Calculation of Work (W)

The work done by the system is given by the formula:

W = -PΔV

where,

P = Constant pressure = 101 kPa

ΔV = Change in volume

Change in volume (ΔV) can be calculated using the formula:

ΔV = V2 - V1

where,

V1 = Initial volume = m/ρ = 200/0.9970 = 200.80 cm³

V2 = Final volume

Final volume (V2) can be calculated using the formula:

V2 = V1 (1 + αΔT)

where,

ΔT = Change in temperature = T2 - T1 = 323 - 298 = 25 K

Substituting the values in the above equation, we get:

V2 = 200.80 (1 + 2.07 * 10^-4 * 25) = 201.79 cm³

Now, substituting the values of P, ΔV and converting the units to joules, we get:

W = -101 * 10^3 * (201.79 * 10^-6 - 200.80 * 10^-6) = -0.203 J

Calculation of Heat (Q)

The heat absorbed by the system is given by the formula:

Q = mCPΔT

where,

m = Mass of water = 200 g

CP = Specific heat capacity of water at constant pressure = 75.30 J/K.mol

ΔT = Change in temperature = T2 - T1 = 323 - 298 = 25 K

Substituting the values in the above equation, we get:

Q = 200/18 * 75.30 * 25 = 20925 J

Calculation of Change in Internal Energy (ΔU)

The change in internal energy (ΔU) is

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