$5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is :

\frac{9}{8} R T_{1}

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\frac{3}{2} R T_{1}

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\frac{15}{8} R T_{1}

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\frac{9}{2} R T_{1}

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Solution

The correct option is A $\frac{9}{8} R T_{1}$
$T V^{γ - 1} = C$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
(gamma for monoatomic gas is $\frac{5}{3}$ )
So, $T_{2} = 4 T_{1}$
Work done on system $= \frac{n R Δ T}{γ - 1} = \frac{9}{8} R T_{1}$

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5.6 litre of helium gas at STP is adiabatically compressed to $0.7$ litre. Taking the initial temperature to be $T_{1}$ , the work done on the gas is:

Q. 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be T, the work done in the process is:

5.6 litre of helium gas at STP is adiabatically compressed to $0.7$ litre. Taking the initial temperature to be $T_{1}$ , the work done on the gas is:

5.6 l i t r e

of helium gas at STP is adiabatically compressed to

0.7 l i t r e

. Taking the initial temperature to be

T_{1}

, the work done in the process is

1.6 L

of helium gas at STP is adiabatically compressed to

0.2 L

. Taking the initial temperature to be

27^{\circ} C

, the work done in the process is
[Take

R = 8.314 J/mol K

]

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5.6L of helium gas at STP is adiabatically compressed to 0.7L. Taking the initial temperature to be T1, the work done in the process is :

The correct option is A 98RT1TVγ−1=CT1(5.6)2/3=T2(0.7)2/3(gamma for monoatomic gas is 53)So, T2=4T1Work done on system =nRΔTγ−1=98RT1

$5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is :

The correct option is A $\frac{9}{8} R T_{1}$
$T V^{γ - 1} = C$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
(gamma for monoatomic gas is $\frac{5}{3}$ )
So, $T_{2} = 4 T_{1}$
Work done on system $= \frac{n R Δ T}{γ - 1} = \frac{9}{8} R T_{1}$