Thermodynamics

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

n1T1+n2T2=nT

(0.1) (200)+ (0.05) (400)= (0.15)T

T=266.67

6.25

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

For expansion in vacuum, workdone, w = 0

For isothermal process,  ΔU=0

According to first law of thermodynamics,

ΔU=q+wq=0

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

(n) λ=5

(n, m) : Integers

2m+12λ=323/25=2m+12n3n=10m+5

N, m are integers.

So,  m=1, n=5, λ=1

m=4n=15λ=13m=7n=25λ=15

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

m (L)=m1S1 (ΔT)

m3.4*105= (200) (4200) (25)

m=61.7

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

E= (I) (t) (A)cos2? θ
(3.3)2π31.43*10-4*12

0.99*10-4

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

U=ncvT

=2*32*R*300=7479J      

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

In case of adiabatic process

Workdone, W=ηR (T2T1)1Y

As, ΔQ=ΔU+W

for adiabatic process :- ΔQ=0

ΔU=W

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

SolZ=101 belong to actinoids

104 belong to group 4

 

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Sol. 2 P b N O 3 2 ? Δ 2 P b O + 2 N O 2 + O 2

2 N O 2 ? N 2 O 4 colourless

N 2 O 4 + 2 N O ? 2 N 2 O 3

Oxidation state in N 2 O 3  is +3

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Sol.

AHH2H=3a2AGH3Al=H23MABC=MMADE=M4

I G = M a 2 12 - M 4 a 2 2 12 + M 4 a 2 3 2 = 11 16 M a 2 12

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