If the pressure of the gas is doubled while the volume is held constant, what happens to its temperature? [section 10.4] if the pressure of the gas is doubled while the volume is held constant, what happens to its temperature? [section 10.4] that is, if p is doubled, t increases by a factor of two. that is, if p is doubled, t increases by a factor of four. that is, if p is doubled, t decreases by a factor of three. that is, if p is doubled, t decreases by a factor of two. that is, if p is doubled, t increases by a factor of three. that is, if p is doubled, t increases by a factor of four.

That is, if P is doubled, T increases by a factor of two.

Explanation:

For an ideal gas, we can write the following equation:

P*V = n* R* T

If we take an initial state, with parameters P₁, V₁, T₁, we can write the equation as follows:

P₁* V₁ = n* R* T₁ (1)

Now, we are told that the same mass of gas, changes its state, doubling its pressure, while the volumen is held constant.

We can use the same equation as above, replacing P₁, V₁, T₁ by the new values P₂, V₂, T₂, where P₂ = 2* P₁, and V₂ = V₁, as follows:

2* P₁* V₁ = n* R* T₂ (2)

Dividing both sides in (2) and (1), we get:

T₂ / T₁ = 2

This means that if the pressure is doubled, temperature increases by a factor of two.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6