sum 28.

BITSAT 2015 Coaching › Thermal physics questions › sum 28.

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iitimcomin ·2009-02-15 03:40:23

SEE SINCE TEMP. DOSNT CHANGE dU =0 ....now Q=W=PdV ......... BUT NOW DA GAS IS IN A CYLINDER...HENCE VOLUME IS A CONSTANT!!!!!!!!!!!!!!!!SO W=0 ........
THEREFORE Q=0!!!!!!!!!!!!!!!!!!!!!(ONLY THEORITICAL!!!!!!)..TELL ME IF IM RONG!!!!!

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nihit.desai ·2009-03-20 22:32:16

Hey iitcoming, i think the answer should be B....
v need to take into account the change in intrnal energy due to change in degrees of freedom...

U_i = N * (5/2RT) (as diatomic has 5 degrees of freedom)

U_f= (2N) * (3/2RT) (as disscoiated gas behaves as monoatmic, having 3 degrees of freedom)

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Rohan Ghosh ·2009-03-21 02:59:35

yes you are right nihit i wonder why the post is pinked

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ANKIT MAHATO ·2009-03-21 03:15:26

why iitcommin's post is pinked ...

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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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sum 28.

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