First law: work, heat, internal energy, enthalpy
Top questions
See all →- The work done during reversible isothermal expansion of one mole of hydrogen gas at 25^°C from pressure of 20 atmosphere to 10 atmosphere is:(Given R = 2.0 cal K^-1 mol^-1) Answer: 3. - 413.14 calories Key idea: for reversible isothermal expansion, w=-2.303 nRTlogP₁/P₂ (work done by the gas is negative in this c... 1
- Which amongst the following options is the correct relation between change in enthalpy and change in internal energy? Answer: 3. Δ H = Δ U + Δ n_g RT Key idea: H = U + PV , so at constant T and P, Δ H = Δ U + PΔ V = Δ U + Δ n_g RT . Here Δ n_g is the change... 1
- Which of the following p-V curve represents maximum work done ? Answer: 3. [graph in figure] Key idea: work done by the gas equals the area under the p-V curve, so find the graph with the largest area. Gr... 1
- The correct option for free expansion of an ideal gas under adiabatic condition is : Answer: 1. q = 0, Δ T = 0 and w = 0 Free expansion means expansion into vacuum, so external pressure is zero and work w=-P_extΔ V=0 . Adiaba... 1
- Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is :[Given that 1 L bar = 100 J] Answer: 4. 30 J Key idea: for expansion against a constant external pressure, the magnitude of work is p_extΔ V , and the question asks for... 1
- A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy Δ U of the gas in joules will be : Answer: 2. -505 J Key idea: well insulated means adiabatic, so q = 0 and Δ U = w . w=-p_extΔ V=-2.5\ atm×(4.50-2.50)\ L=-5\ L atm . Taking 1... 1