In a thermodynamic process two moles of a monotomic ideal gas obeys p∝V2. If temperature of the gas increases from 300K to 400K, then find work done by the gas (where R= universal gas constant)

pV2= constant(as p∝V-2 )Comparing with pVx= constant, we havex=2W=nRΔT1-xSubstituting the values, we haveW=(2)(R)(400-300)1-2=-200R

In a thermodynamic process two moles of a monotomic ideal gas obeys p∝V2. If temperature of the gas increases from 300K to 400K, then find work done by the gas (where R= universal gas constant)

pV2= constant(as p∝V-2 )Comparing with pVx= constant, we havex=2W=nRΔT1-xSubstituting the values, we haveW=(2)(R)(400-300)1-2=-200R

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In a thermodynamic process two moles of a monotomic ideal gas obeys $p \propto V^{2}$ . If temperature of the gas increases from $300 K$ to $400 K$ , then find work done by the gas (where $R =$ universal gas constant)

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$- 100 R$

$200 R$

$- 20 \times 0 n$

answer is B.

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Detailed Solution

$p V^{2} =$ constant

(as $p \propto V^{- 2}$ )

Comparing with $p V^{x} =$ constant, we have

$x = 2$

$W = \frac{n R Δ T}{1 - x}$

Substituting the values, we have

$W = \frac{(2) (R) (400 - 300)}{1 - 2}$

$= - 200 R$

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In a thermodynamic process two moles of a monotomic ideal gas obeys p∝V2. If temperature of the gas increases from 300K to 400K, then find work done by the gas (where R= universal gas constant)

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In a thermodynamic process two moles of a monotomic ideal gas obeys $p \propto V^{2}$ . If temperature of the gas increases from $300 K$ to $400 K$ , then find work done by the gas (where $R =$ universal gas constant)