Question

1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature is 0°C. Determine the work done during this cooling.

Answer 1

Answer:

the work done during this cooling is −28.7 kJ

Explanation:

Given data

mass (m) = 1 kg

r = 287 J/kg-K

pressure ( p) = 50 kPa

temperature (T) = 100°C = ( 100 +273 ) = 373 K

to find out

the work done during this cooling

Solution

we know the first law of thermodynamics

pv = mRT     ....................1

here put value of p, m R and T and get volume v(a) when it initial stage in equation 1

50 v(a) = 1 × 0.287  × 373

v(a) = 107.051 / 50

v(a) = 2.1410 m³    .......................2

now we find out volume when temperature is  0°C

so put  put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1

50 v(b) = 1 × 0.287  × 273

v(a) = 78.351 / 50

v(a) = 1.5670 m³    .......................3

by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.

work done = $\int\limits^a_b {p} \, dv$

integrate it and we get

work done = p ( v(b) - v(a)  ) ................4

put the value p and v(a) and v(b) in equation 4 and we get

work done = p ( v(b) - v(a)  )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 (−0.574)

work done = −28.7 kJ

here we can see work done is negative so its mean work done opposite in direction of inside air

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