Voltmeter forming a closed circuit

A battery is connected to a 10Ω resistor as shown in Figure 2. The emf (electromotive force) of the battery is 6.0 V.

When the switch is open the voltmeter reads 6.0 V and when it is closed it reads 5.8V. Explain why the readings are different.

when switch is closed a current flows (through the battery)
hence a pd/lost volts develops across the internal resistance

But why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance, without having the other loop's switch closed?

-

why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance

It does, but the voltmeter probably has an impedance of 10,000,000 ohms - as most multimeters do - so the current flowing is negligibly small (0.6 microAmps) and therefore the voltage drop due to battery internal resistance is also negligible.

4 x Alkaline AA has resistance maybe 4 x 300 milliohms. At 0.6 microAmps that's a voltage drop of 0.72 microvolts (if I did my sums right - please check)

-

The voltmeter does indeed draw current, but very small. Its internal resistance should be high, unless it's a very cheap voltmeter.

Putting the voltmeter's internal resistance, let us say 1MOhm just to make up a value, in series with the battery's internal resistance of a fraction of one Ohm, makes a voltage divider providing practically the entire battery voltage, the drop begin too small to notice. Your 10 Ohm resistor, on the other hand, is only one or two orders of magnitude larger than the battery internal resistance, and so produces a noticeable voltage drop.

-