a given mass of an ideal gas law occupies a volume V of a temperature T under a pleasure p, if the pressure is increased to 2 p and temperature 1/2 T what is the percentage change in the volume of the gas​

75%

Explanation:

We'll begin by calculating the final volume of the gas. This can be obtained as follow:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

Final pressure (P₂) = 2P

Final temperature (T₂) = ½T

Final volume (V₂) =?

P₁V₁/T₁ = P₂V₂/T₂

PV/T = 2P × V₂ / ½T

Cross multiply

T × 2P × V₂ = PV × ½T

T × 2P × V₂ = PV × ½T

Divide both side by T × 2P

V₂ = (PV × ½T) / T × 2P

V₂ = ¼V

Next, we shall determine the absolute change in the volume of the gas. This can be obtained as follow:

Initial volume (V₁) = V

Final volume (V₂) = ¼V

Absolute change in volume |ΔV| =?

|ΔV| = |V₂ – V₁|

|ΔV| = |¼V – V|

|ΔV| = |0.25V – V|

|ΔV| = 0.75V

Finally, we shall determine percentage change in the volume of the gas. This can be obtained as follow:

Initial volume (V₁) = V

Absolute change in volume |ΔV| = 0.75V

Percentage change =?

Percentage change = |ΔV| / V × 100

Percentage change = 0.75V / V × 100

Percentage change = 0.75 × 100

Percentage change = 75%

Thus, the percentage change in the volume of the gas is 75%

Explanation:

I think is B