The temperature of n moles of an ideal gas is increased from T to 4T through a process for which pressure where a is constant. The work done by gas is:

nRT

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4nRT

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2nRT

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6nRT

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Solution

The correct option is D
6nRT

PV =nRT
$V = \frac{n R T}{P} V = \frac{n R T}{a} T = \frac{n R T^{2}}{a} ∴ P = \frac{a}{T} d V = \frac{2 n R T}{a} d t d W = \int_{T}^{4 T} P d V = \int_{T}^{4 T} P . \frac{2 n R T}{a} d t = \int_{T}^{4 T} \frac{a}{T} . \frac{2 n R T}{a} d t W = 6 n R T$

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Similar questions

P = a T^{- 1}

T t o 4 T

p = a T^{- 1}

P = α T^{1 / 2}

α

50 K

T = K V^{2}

T_{0}

4 T_{0}

T_{1} t o T_{2}

x R (T_{2} - T_{1})

x

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