A cylinder of ideal gas is closed by an 8kg movable piston of area `60cm^2`. The atmospheric pressure is 100kPa. When the gas is heated form `30^@C` to `100^@C`, the piston rises 20 cm. The piston is then fastened in the place and the gas is cooled back to `30^@C`. If `DeltaQ_1` is the heat added to the gas during heating and `DeltaQ_2` is the heat lost during cooling, find the difference.

Options

A. zero
B. `136 J`
C. `- 136 J`
D. `- 68 J`

Answer

B

Explanation

b. The gas pressure
`= ("weight of piston")/("Area of cross-section") + atm`. pressure
`= (8 xx 9.8)/(60 xx 10^(-4)) + 1.00 xx 10^(5) N//m^(2)`
`= 1.13 xx 10^(5) N//m^(2)`
During the heat pressure, the internal energy is changed by `Delta U_(1)` and work `Delta W_(1)` is done.
Therefore, `Delta Q_(1) = Delta U_(1) + Delta W_(1) = Delta U_(1) + PdV`
`Delta_(1) + (1.13 xx 10^(5)) (0.02 xx 60 xx 10^(-4))`
`Delta U_(1) + 136 J`
During the cooling process, no work is done as volume is constant, `Delta W = 0`
Hence, `Delta Q_(2) = Delta U_(2)`, But `Delta U_(2)` is negative as the temperature decreases, and since the gas returns to its original temperature, `Delta U_(2) = - Delta U_(1)`
Hence
`[Delta Q_(1) - | Delta Q_(2) |] = (Delta U_(1) + 136 - Delta U_(1)) = 136 J`