A mass of gas was compressed without friction at constant pressure of 100 kPa from $0.3 m^{3}$ of initial state to the $0.15 m^{3}$ of final state. There was a transfer of 48 kJ heat from the gas during the process. The decrease in internal energy of the gas is

33 kJ

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-33 kJ

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28.5 kJ

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42 kJ

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Solution

The correct option is A 33 kJ
$Initial state : P_{1} = 100 k P a, V_{1} = 0.3 m^{3}$

$Final state : P_{2} = 100 k P a, V_{2} = 0.15 m^{3},$

$Q = - 48 k J$

Now,

$Q_{1 - 2} = Δ U + W_{1 - 2} - - - - (1)$

$W_{1 - 2} = \int_{v_{1}}^{v_{2}} P d V$

$= - 100 \times (0.3 - 0.15) = - 15 k J$

From eq. (1)

$U_{2} - U_{1} = - 48 + 15 = - 33 k J$

$Decrease in internal energy = 33 k J$

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A mass of gas was compressed without friction at constant pressure of 100 kPa from 0.3m3 of initial state to the 0.15m3 of final state. There was a transfer of 48 kJ heat from the gas during the process. The decrease in internal energy of the gas is

The correct option is A 33 kJInitial state :P1=100 kPa, V1=0.3 m3 Final state :P2=100 kPa,V2=0.15m3, Q=−48kJ Now, Q1−2=ΔU+W1−2−−−−(1) W1−2=∫v2v1PdV =−100×(0.3−0.15)=−15 kJ From eq. (1) U2−U1=−48+15=−33kJ Decrease in internal energy=33kJ

A mass of gas was compressed without friction at constant pressure of 100 kPa from $0.3 m^{3}$ of initial state to the $0.15 m^{3}$ of final state. There was a transfer of 48 kJ heat from the gas during the process. The decrease in internal energy of the gas is