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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.
Related Questions: - Primary,secondary and tertiary amines may be separated by using
- Formula of hexaaquamanganese(II)phosphate is
- Which of the following exists as zwitter ion
- The equation ΔG = ΔH – TΔS is for
- A secondary amine could be prepared readily from the starting material
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Primary,secondary and tertiary amines may be separated by using
- Formula of hexaaquamanganese(II)phosphate is
- Which of the following exists as zwitter ion
- The equation ΔG = ΔH – TΔS is for
- A secondary amine could be prepared readily from the starting material
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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