Thermodynamics
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In the above diagram if a systems from initial state A (P1,V1,T1) goes to final state C (P2 ,V2 ,T1) from 3 different paths: Path 1 is A to C, Path 2 is A to B to C, Path 3 is A to D to C If 5 kJ of work is done by system in Path A to C then how much heat is transferred during Path A to D is:
In the above diagram if a systems from initial state A (P1,V1,T1) goes to final state C (P2 ,V2 ,T1) from 3 different paths: Path 1 is A to C, Path 2 is A to B to C, Path 3 is A to D to C If 5 kJ of work is done by system in Path A to C then how much heat is transferred during Path A to D is:
(0) Path A to D is adiabatic. &nbs
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The temperature of equals masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. Temperature of mixture when x and z are mixed will be:
The temperature of equals masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. Temperature of mixture when x and z are mixed will be:
12gm of gas occupy a volume of 4 × 10-3 m3 at a temperature of 7°C. After the gas is heated at constant pressure its density becomes 6 × 10-4gm/cm3. What is the temperature to which the gas was heated (in kelvin)?
12gm of gas occupy a volume of 4 × 10-3 m3 at a temperature of 7°C. After the gas is heated at constant pressure its density becomes 6 × 10-4gm/cm3. What is the temperature to which the gas was heated (in kelvin)?
P = (RT/M? )ρ ⇒ P/ρT = constant
⇒ (12)/ (4 × 10? ³ × 10? ) × 280 = 6 × 10? × T ⇒ T = 1400K
Upvote The molar specific heat Cv of a diatomic gas is (5/2)R when we treat its molecules to be rigid having no vibrations. But this assumption is not always valid especially at high temperatures. In that case, molecules are treated as non-rigid having vibrational energy also. Cv for diatomic gas with non-rigid gas molecules will be:
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The molar specific heat Cv of a diatomic gas is (5/2)R when we treat its molecules to be rigid having no vibrations. But this assumption is not always valid especially at high temperatures. In that case, molecules are treated as non-rigid having vibrational energy also. Cv for diatomic gas with non-rigid gas molecules will be:
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Vibrational energy of a non-rigid gas molecule is K? T so, total energy = (5/2)K? T + K? T = (7/2)K? T
∴ C? = (7/2)R
Upvote The equilibrium constant Kp for a homogenous gaseous reaction is 10-3, the standard Gibb's free energy change (ΔG°) in cal mol-1 for the reaction at 27°C (using R = 2 cal K-1 mol-1) is :
The equilibrium constant Kp for a homogenous gaseous reaction is 10-3, the standard Gibb's free energy change (ΔG°) in cal mol-1 for the reaction at 27°C (using R = 2 cal K-1 mol-1) is :
ΔG° = – 2.303RT log Kp = – 2.303× 2 × 300log10? ³
= – 2.303× 2 × 300 × (– 3) = 4145.4cal mol? ¹
Upvote ΔH_f° of CO₂(g), CO(g), N₂O(g) and NO₂(g) in kJ mol⁻¹ respectively is -393, -110, 81 and 34. Calculate the ΔH in kJ for the following reaction :
2NO₂(g) + 3CO(g) → N₂O(g) + 3CO₂(g).
ΔH_f° of CO₂(g), CO(g), N₂O(g) and NO₂(g) in kJ mol⁻¹ respectively is -393, -110, 81 and 34. Calculate the ΔH in kJ for the following reaction :
2NO₂(g) + 3CO(g) → N₂O(g) + 3CO₂(g).
ΔH_rxn = ΔH_f (N? O, g) + 3ΔH_f (CO? , g) – (2ΔH_f NO? , g) – 3ΔH_f (CO, g)
= 81 + 3× (– 393) – 2 × 34 – 3 (–110)
= 81 – 1179 – 68 + 330 = – 836 kJ
Upvote One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process, which one of following statement is correct?
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One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process, which one of following statement is correct?
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One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process ΔG = 0. As phase transformation of water is an equilibrium process and at equilibrium, free energy change is always zero.
Upvote A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then :
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A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then :
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2CO (g) + O? (g) → 2CO? (g)
Δn_g = 2 – (2 + 1) = -1
ΔH = ΔE + Δn_g RT or ΔH = ΔE – 1RT
i.e. ΔH < ΔE
Upvote A carnot engine works between temperatures 327°C and 27°C. If the engine takes 1600 J of heat from the higher temperature reservoir, the work done by the engine per cycle (in Joule) is equal to ______.
A carnot engine works between temperatures 327°C and 27°C. If the engine takes 1600 J of heat from the higher temperature reservoir, the work done by the engine per cycle (in Joule) is equal to ______.
(Q? - Q? )/Q? = (T? - T? )/T? ; Q = (300/600) × 1600 = 800 J
work done = Q? - Q? = 1600 – 800 = 800J
Upvote If the molar specific heat at constant pressure for a polyatomic non-linear gas is x and the molar specific heat at constant volume for a diatomic gas is y, find the value of x y
If the molar specific heat at constant pressure for a polyatomic non-linear gas is x and the molar specific heat at constant volume for a diatomic gas is y, find the value of x y
x = 8R/2 ; y = 5R/2 ; x/y = 8/5 = 1.6
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