40 g of Argon is heated from 40⁰C to 100⁰C (R=2 cal/mole). Heat absorbed at constant

40 G Of Argon Is Heated From 40c To 100c Physics Question

40 g of Argon is heated from 40⁰C to 100⁰C (R=2 cal/mole). Heat absorbed at constant volume is

Options

(a) 100 cal
(b) 80 cal
(c) 180 cal
(d) 120 cal

Correct Answer:

180 cal

Explanation:

Heat absorbed at constant volume = nCvdT
Now argon is monoatomic Cv = (3/2) R
Number of moles = 4/0/40 = 1
.·. Q = 1 × (3/2) × 2 × (100-40) = 3 × 60 = 180 cal.

Related Questions:

  1. Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
  2. The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
  3. A condenser of capacity C is charged to a potential difference of V₁. The plates
  4. Two identical thin plano-convex glass lenses (refractive index 1.5) each having
  5. Which of the following is the function of the step-up transformer?

Topics: Thermodynamics (179)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*