The pressure in the gas in a flow device changes according to the relation $P = - 1000 V + C$ where P is in kPa and v in $m^{3} / k g$ initially the pressure is 100 kPa and volume is 0.01 $m^{3} / k g$ . What is the work done if the final pressure is 50 kPa ?

1.75 kJ

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3.25 kJ

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3.75 kJ

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4.5kJ

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Solution

The correct option is A 1.75 kJ
$P_{1} = 100 k P a v_{1} = 0.01 m^{3} / k g$

$P_{2} = 50 k P a$

$P = - 1000 v + C$

$100 = - 1000 \times 0.01 + C$

$110 = C$

$P = - 1000 v + 110$

$\frac{100 - C}{- 1000} = v$

$\frac{100 - C}{- 1000} = + 0.01 \Rightarrow C = 110$

$\frac{P - 110}{- 1000} = v$

$Work done = \int - v d P = \frac{P - 110}{1000} d P$

$\frac{1}{1000} [P^{2} - 110 P] = - \frac{1}{1000} {\frac{(50^{2} - 100^{2})}{2} - 110 [50 - 100]} = 1.75 k J$

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The pressure in the gas in a flow device changes according to the relation P=−1000 V+C where P is in kPa and v in m3/kg initially the pressure is 100 kPa and volume is 0.01 m3/kg. What is the work done if the final pressure is 50 kPa ?

The correct option is A 1.75 kJP1=100 kPa v1=0.01 m3/kg P2=50 kPa P=−1000 v+C 100=−1000×0.01+C 110=C P=−1000v+110 100−C−1000=v 100−C−1000=+0.01⇒C=110 P−110−1000=v Work done =∫−vdP=P−1101000dP 11000[P2−110P]=−11000{(502−1002)2−110[50−100]}=1.75kJ

The pressure in the gas in a flow device changes according to the relation $P = - 1000 V + C$ where P is in kPa and v in $m^{3} / k g$ initially the pressure is 100 kPa and volume is 0.01 $m^{3} / k g$ . What is the work done if the final pressure is 50 kPa ?