A piston–cylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant pressure until T 2 = 3 4 T 1 . Then the piston is held stationary while the gas is further cooled to T 3 = 1 2 T 1 , where all temperatures are in K. 1. The ratio of the final volume to the initial volume of the gas is (a) 0.25 (b) 0.50 (c) 0.67 (d) 0.75 (e) 1.0 2. The work done on the gas by the piston is (a) RT 1 /4 (b) c v T 1 /2 (c) c p T 1 /2 (d) ( c v + c p ) T 1 /4 (e) c v ( T 1 + T 2 )/2 3. The total heat transferred from the gas is (a) RT 1 /4 (b) c v T 1 /2 (c) c p T 1 /2 (d) ( c v + c p ) T 1 /4 (e) c v ( T 1 + T 3 )/2
A piston–cylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant pressure until T 2 = 3 4 T 1 . Then the piston is held stationary while the gas is further cooled to T 3 = 1 2 T 1 , where all temperatures are in K. 1. The ratio of the final volume to the initial volume of the gas is (a) 0.25 (b) 0.50 (c) 0.67 (d) 0.75 (e) 1.0 2. The work done on the gas by the piston is (a) RT 1 /4 (b) c v T 1 /2 (c) c p T 1 /2 (d) ( c v + c p ) T 1 /4 (e) c v ( T 1 + T 2 )/2 3. The total heat transferred from the gas is (a) RT 1 /4 (b) c v T 1 /2 (c) c p T 1 /2 (d) ( c v + c p ) T 1 /4 (e) c v ( T 1 + T 3 )/2
Solution Summary: The author explains the ratio of the final volume to the initial volume in Equation (I) and the work done by the piston cylinder device.
A piston–cylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant pressure until
T
2
=
3
4
T
1
. Then the piston is held stationary while the gas is further cooled to
T
3
=
1
2
T
1
, where all temperatures are in K.
1. The ratio of the final volume to the initial volume of the gas is
36
2) Use the method of MEMBERS to determine the true magnitude and
direction of the forces in members1 and 2 of the frame shown below
in Fig 3.2.
300lbs/ft
member-1
member-2
30°
Fig 3.2.
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5670 mm
The apartment in the ground floor of three floors building in Fig. in Baghdad city. The details of
walls, roof, windows and door are shown. The window is a double glazing and air space thickness
is 1.3cm Poorly Fitted-with Storm Sash with wood strip and storm window of 0.6 cm glass
thickness. The thickness of door is 2.5 cm. The door is Poor Installation. There are two peoples
in each room. The height of room is 280 cm. assume the indoor design conditions are 25°C DBT
and 50 RH, and moisture content of 8 gw/kga. The moisture content of outdoor is 10.5 gw/kga.
Calculate heat gain for living room :
الشقة في الطابق الأرضي من مبنى ثلاثة طوابق في مدينة بغداد يظهر في مخطط الشقة تفاصيل الجدران والسقف
والنوافذ والباب. النافذة عبارة عن زجاج مزدوج وسمك الفراغ الهوائي 1.3 سم ضعيف الاحكام مع ساتر حماية مع إطار
خشبي والنافذة بسماكة زجاج 0.6 سم سماكة الباب 2.5 سم. الباب هو تركيب ضعيف هناك شخصان في كل غرفة.
ارتفاع الغرفة 280 سم. افترض أن ظروف التصميم الداخلي هي DBT25 و R50 ، ومحتوى الرطوبة 8…
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