Thermodynamics

Question

A gas expands from 3 dm³ to 5 dm³ against a constant external pressure of 3atm. The work done during expansion is used to heat 10mol of water at 290 K. Calculate the final temperature of water. Sp. Heat of water = 4.184 J g^-1 K^-1

Answer 1

 

I will set this up for you and you can do the number crunching.

 

At constant pressure, 

 

W = PΔV = P(V_f - V_i)

V_f = final volume of gas (given)

V_i = initial volume of gas (given)

c = specific heat of water (given)

n = number of moles of water (given)

M = molecular weight of water (add the atomic weights of 2H and O, or look up in tables)

m = mass of water = nM

Q = heat

T_i = initial temperature of water (given)

T_f = final temperature of water = ?

 

Q = mcΔT = nMc(T_f - T_i)

 

Assume all the work is converted to heat.

 

W = Q

 

P(V_f - V_i) = nMc(T_f - T_i)

 

T_f = T_i + [P(V_f - V_i)]/(nMc)

 

Fill in the numbers in compatible units and crunch out the answer.

Answer 2

Work done = P ΔV = 2 dm³ X 3 atm = 6.06X10²J = ρcΔTv_water⁼1000X4.184*18X10^-3XΔT; Therefore, ΔT=8.04 K