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Standard XII

Physics

Work

The pressure ...

Question

The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa. 50 cc.

(a) Calculate the work done by the gas.

(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas ?

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Solution

$P_{1} = 10 k Pa$

$= 10 \times 10^{3} Pa$

$P_{2} = 50 \times 10^{3} Pa$

$V_{1} = 200 CC, V_{2} = 50 CC$

(i) work done on the gas

$= [(\frac{1}{2}) (10 + 50) \times 10^{3} \times (50 - 200) \times 10^{- 6}]$

$= - 4.5 J$

(ii) $Δ Q = 0$ ,

So, $Δ U = - Δ W = 4.5 J$

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The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa. 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas ?

P1=10 k Pa =10×103 Pa P2=50×103 Pa V1=200 CC, V2=50 CC (i) work done on the gas =[(12)(10+50)×103×(50−200)×10−6] =−4.5 J (ii) ΔQ=0, So, ΔU=−ΔW=4.5 J

The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa. 50 cc.

(a) Calculate the work done by the gas.

(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas ?

$P_{1} = 10 k Pa$

$= 10 \times 10^{3} Pa$

$P_{2} = 50 \times 10^{3} Pa$

$V_{1} = 200 CC, V_{2} = 50 CC$

(i) work done on the gas

$= [(\frac{1}{2}) (10 + 50) \times 10^{3} \times (50 - 200) \times 10^{- 6}]$

$= - 4.5 J$

(ii) $Δ Q = 0$ ,

So, $Δ U = - Δ W = 4.5 J$