Answer
a) $0.125$ $L$.
b) $56.98$ $J$ work is done on the gas.
c) Internal energy has decreased.
d) Heat left the system.
Work Step by Step
a) The number of gas particles is constant. By ideal gas law,
$PV\propto T$, where symbols have the usual meaning.
At a:
$P=1.50$ $atm$, $V=0.5$ $L$, temperature $=T_o$(say)
At b:
$P=1.50$ $atm$, temperature $=\frac{1}{4}T_o$
Pressure is constant. Thus $V\propto T$.
Thus, by the ideal gas law, volume at b $=\frac{1}{4}0.5$ $L$ $=0.125$ $L$.
b) Difference in volume from a to b $=(0.500-0.125)$ $L$
$=0.375$ $L$
Pressure = 1.50 atm
Work done $=(1.50\times 0.375)$ $ L-atm$ $=0.5625$ $L-atm$
$1$ $L-atm=101.3$ $J$.
Thus, work done =$ (0.5625\times 101.3)$ $J$ $\simeq 56.98$ $J$.
Volume has decreased from a to b. So work was done on the gas.
c) For ideal gas, internal energy $U\propto T$.
From a to b, temperature has decreased. Thus, internal energy has decreased.
d) Work done on gas must increase its internal energy. But as $U$ has decreased overall, energy must also leave the system as heat simultaneously for consistency.
