One mole of an ideal monatomic gas at temperature T 0 expands solely according to the law P V - constant- If the final temperature is 2 T 0 - heat supplied to the gas is-

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Entropy Change in Irreversible Process

One mole of a...

Question

One mole of an ideal monatomic gas at temperature $T_{0}$ expands solely according to the law $\frac{P}{V}$ = constant. If the final temperature is $2 T_{0}$ , heat supplied to the gas is:

2 R T_{0}

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R T_{0}

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\frac{3}{2} R T_{0}

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\frac{1}{2} R T_{0}

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Solution

The correct option is A $2 R T_{0}$
In a process $P V^{x} =$ constant, molar heat capacity is given by C $= \frac{R}{γ + 1} + \frac{R}{1 - x}$

As the process is $\frac{P}{V} =$ constant,

i.e., $P V^{- 1} =$ constant, therefore, $x = - 1$ .

For an ideal monatomic gas, $γ = \frac{5}{3}$

$∴ C = \frac{R}{\frac{5}{3} - 1} + \frac{R}{1 - (- 1)} = \frac{3}{2} R + \frac{R}{2} = 2 R$

$Δ Q = n C (Δ T) = 1 (2 R) (2 T_{0} - T_{0}) = 2 R T_{0}$

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One mole of an ideal monatomic gas at temperature T0 expands solely according to the law PV = constant. If the final temperature is 2T0 , heat supplied to the gas is:

The correct option is A 2RT0In a process PVx=constant, molar heat capacity is given by C=Rγ+1+R1−xAs the process is PV= constant,i.e., PV−1=constant, therefore, x=−1.For an ideal monatomic gas, γ=53∴C=R53−1+R1−(−1)=32R+R2=2RΔQ=nC(ΔT)=1(2R)(2T0−T0)=2RT0

One mole of an ideal monatomic gas at temperature $T_{0}$ expands solely according to the law $\frac{P}{V}$ = constant. If the final temperature is $2 T_{0}$ , heat supplied to the gas is: