If the pressure of a confined gas doubles and its volume decreases by a factor of four, how will its kelvin temperature change?

Temperature halves or decreases by a factor of 2

The relationship between pressure, volume, and temperature can be expressed as;

 PV/T

and we can compare one state to another by this expression;

P1V1/T1 = P2V2/T2

in this case, we can say P1, V1 and T1 are all equal to 1 and have the pressure doubled, such that P2 = 2 and the volume decreases by a factor of four, so that V = 1/4; then T2 will be equal to;

(1)(1)/1 = (2)(1/4)/T2

  1 = 0.5/T2

T2 = 0.5 or 1/2

Temperature will therefore half or decrease by a factor of 2

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