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When 1 mol of gas is heated at constant volume, temperature is raised from 298 to 308 K. Heat supplied to the gas is 500J. Then which statement is correct
Options
(a) q = w = 500 J,ΔU = 0
(b) q = ΔU = 500 J,W = 0
(c) q = w = 500 J ,ΔU = 0
(d) ΔU = 0, q = w = -500J
Correct Answer:
q = ΔU = 500 J,W = 0
Explanation:
We known that ΔH = ΔE + PV
ΔH = ΔE + P Δ V + V Δ P =0
When ΔV = 0; w = 0. Therefore ΔH = ΔE + PΔV
ΔH = ΔE + 0 or ΔH = ΔE.
As ΔE = q + w , ΔE = q.
In the present problem, ΔH = 500J,
ΔV = ΔE = 500 J,q = 500 J, w = 0.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ionisation potential for hydrogen atom is 13.6eV, the ionization potential for He⁺
- Phenol on reaction with bromine water would give
- The volume of water to be added to N/2 HCl to prepare 500 cm³ of N/10 solution is
- As we go from left to right in a period of the periodic table, gram atomic volume
- If 1 mole of an ideal gas expands isothermally at 37⁰C from 15 litres
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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