Question

An ideal gas undergoing expansion in vacuum shows:
(a) $\vartriangle E=0$
(b) $W=0$
(c) $Q=0$
(d) All of these

Answer 1

Hint: We have been given the ideal gas in undergoing expansion in vacuum. And we have to find what is the value of different parameters and change these parameters under this condition. For that, we shall first acknowledge the properties of vacuum and then find out what effect does vacuum have on these different parameters.

Complete answer:
A vacuum is a space which is completely devoid of matter. The pressure in vacuum is so very low that it almost tends to zero. This means that even if we release a few particles in vacuum, they will have no effect whatsoever in any process which is being carried out there. It is a condition well below normal atmospheric pressure and is measured in units of Pascal (also a unit of pressure).
Now, for a gas undergoing expansion in vacuum, we can say that it is a case of free expansion. This means that expansion under the effect of no external pressure.
 Also, the formula for work done by an ideal gas is given by:
$\Rightarrow W={{P}_{external}}\int{dV}$
Here, the value of external pressure is zero, which means the net work done is equal to:
$\Rightarrow W=0$
This implies that the second option is correct.
Now, for energy interaction of a system with a surrounding, there must be a surrounding. Since, vacuum is just empty space, it has no surrounding. Therefore, no heat is exchanged by the gas. Hence, we can say that $Q=0$ .
This implies the third option is also correct.
Now, from the first law of thermodynamics. We can write as:
$\Rightarrow Q=U+\vartriangle W$
Here, $Q=\vartriangle W=0$
Putting these values in the above expression, we get:
$\Rightarrow U=0$
Hence, the first option is also correct.
Therefore, we see that the values of $U,\vartriangle W,Q$ come out to be zero.
This means all the three options are correct.

Hence, option (d)All of these, is the correct option.

Note:
In theoretical or conceptual questions like these, one should be very quick to solve them as it will help save a bunch of time. Also, in free expansion, since there is no change in internal energy of the body, this means that even the temperature has remained constant over time.