Question

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83.8 J of heat is transferred to the steam, causing the steam temperature and steam volume to rise at constant pressure (125 kPa). The specific enthalpy (in J/mol) of steam at 125 kPa varies with temperature as 34980+35.5T (where temperature T is in K). Taking the steam as the system, formulate the first law of thermodynamics. Then calculate
a. find the steam temperature in K
b. final cylinder volume
c. work is done by the steam
d. change in internal energy of the steam in J

Answer 1

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785$cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$⋅Pa⋅$K^{-1}$⋅$mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5$T_{1}$ - (34980 + 35.5$T_{o}$)]

83.8 = (0.03 × 35.5) ($T_{1}$ - 400K)

83.8 = 1.065 ($T_{1}$  - 400)

78.69 = ($T_{1}$  - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$/$T_{1}$ = $V_{o}$/$T_{o}$

$V_{1}$  =  $V_{o}$$T_{1}$/$T_{o}$

$V_{1}$   = (785 × 478.69)/400

$V_{1}$   = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P($V_{1}$- $V_{o}$) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J