one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is (1 L atm = 101.3 J)A. `5.763`B. `1.013`C. `-1.013`D. `-5.763`

one mole of an ideal gas at 300k in thermal contact with surroundings

1.	one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is (1 L atm = 101.3 J)A. `5.763`B. `1.013`C. `-1.013`D. `-5.763`
Answer» Correct Answer - C In Isothermal process, `DeltaU=0`, then `q_("irr")= - W_("irr")` `q_("irr")= - (-P_(ect) DeltaV)` `= 3` L atm `=3xx101.3 J=303.9 J` `DeltaS_(surr)=q_(surr)/T=(-q_(sys))/T=(-303.9 J)/(300 K)=-1.013 J K^(-1)` `C(s)+O_(2)(g) rarrCO_(2)(g), DeltaH=-393.5 kJ mol^(-1)` ...(i)

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one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is (1 L atm = 101.3 J)A. `5.763`B. `1.013`C. `-1.013`D. `-5.763`

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