Q.

We make helium gas of 2 g to go through the thermal cycle shown in the figure. The lowest temperature of the gas in the cycle is -17C the highest is 127C and the temperature is equal at points A and B. [R=Universal gas constant]

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a

the temperature at A & B is 328 K

b

net work done by the gas during a cycle is 32 R

c

the temperature of A & B is 320 K

d

net work done by the gas during a cycle is 8 R

answer is C, D.

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Detailed Solution

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Process DA is isochoric, P0256=PT1            … (1)

              CB is isochoric, P0400=PT1            … (2)

from equation (1) × equation (2)
we get T1=320 K

WAB=ηRT             =24R400-320=40R WAB=ηRTD-TB=-32R Wnet=8R

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