Fig shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

Answer

B, C

Explanation

Given ,`n = 2moles`.`
`DeltaV = 0`
`In ab and bc`
`Hence DeltaW = DeltaQ`
`DeltaW =DeltaW_AB+DeltaW_CD`
` = nRT_1 In((2V_0)/(V_0)+nRT_2 In((V_0)/(2V_0))`
`nRxx2.303xxlog 2xx(500-300)`
`= 2xx8.314xx2.303xx0.301xx200`
` = 2305.31J