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Thermodynamic Process Physics Thermodynamics Physics Class 11th

The temperature of 3.00 mol of an ideal diatomic gas in increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of work done by the gas is $\frac{x}{10} .$ Then the value of x (round off to the nearest integer) is________.

(Given R = 8.31 J mol^-1K^-1)

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alok kumar singh | Contributor-Level 10

2 months ago

Since process is isochoric

So $Δ U = n C_{v} Δ T$

$Δ U = n (\frac{5}{2} R) Δ T - - (i) [C_{V} = \frac{5}{2} R]$

And external work

$Δ W = n R Δ T - - (i i)$

$\frac{5}{2} = \frac{x}{10} \Rightarrow x = 25.00$

Since process is isochoricSo <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816002803"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mi>n</mi> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>v</mi> </mrow> </msub> <mi>Δ</mi> <mi>T</mi> </mrow> </math> <math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mi>n</mi> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>R</mi> </mrow> <mo>)</mo> </mrow> <mi>Δ</mi> <mi>T</mi> <mo>−</mo> <mo>−</mo> <mo stretchy="false"> (</mo> <mi>i</mi> <mo stretchy="false">)</mo> <mtext> </mtext> <mrow> <mo> [</mo> <mrow> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>V</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>R</mi> </mrow> <mo>]</mo> </mrow> </mrow> </math>    <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816002804"> </o:OLEObject></xml><! [endif]-> And external work <math> <mrow> <mi>Δ</mi> <mi>W</mi> <mo>=</mo> <mi>n</mi> <mi>R</mi> <mi>Δ</mi> <mi>T</mi> <mtext> </mtext> <mo>−</mo> <mo>−</mo> <mrow> <mo> (</mo> <mrow> <mi>i</mi> <mi>i</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math>    <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816002805"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> <mo>⇒</mo> <mi>x</mi> <mo>=</mo> <mn>2</mn> <mn>5</mn> <mo>.</mo> <mn>0</mn> <mn>0</mn> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1816002806"> </o:OLEObject></xml><! [endif]->

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A calorimeter of water equivalent $20 g$ contains $180 g$ of water at $25^{?} C$ , ' $m$ ' grams of steam at $100^{?} C$ is mixed in it till the temperature of the mixture is $31^{?} C$ . The value of

alok kumar singh

Heat lost by steam $=$ Heat gained by water and calorimeter.

$\begin{array}{r} m \times 540 + m \times 1 \times (100 - 31) = 200 \\ 540 m + 69 m = 1200 \\ m = \frac{1200}{609} \approx 2 \end{array}$

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alok kumar singh

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alok kumar singh

N_a -> Avogadro’s Number

$J β^{2} \equiv [μ k R] \Rightarrow β^{2} \equiv [k R] \Rightarrow β \equiv [k] \equiv [M L^{2} T^{- 2} K^{- 1}]$

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-> $S \equiv α^{2} β \Rightarrow [M L^{2} T^{- 2} K^{- 1}] \equiv α^{2} [M L^{2} T^{- 2} K^{- 1}] \Rightarrow α \equiv [M^{0} L^{2} T^{0} K^{0}]$ Dimensionless

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Heat absorbed in cyclic process = Work done = 100? Joule

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alok kumar singh

Since process is isochoric, so

$Q = n C_{v} Δ T = n (\frac{f}{2} R) Δ T = 4 (\frac{5}{2} \times R) (50) = 500 R$

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