If the pressure of a gas sample is quadrupled and the absolute temperature is doubled, by what factor does the volume of the sample change

The new volume of the sample is halved.

Explanation:

Data obtained from the question include the following:

Initial volume (V1) = V

Initial temperature (T1) = T

Initial pressure (P1) = P

Final pressure (P2) = quadrupled = 4P

Final temperature (T2) = doubled = 2T

Final volume (V2) =?

Thus, we can obtain the new volume of the same by using the combined gas equation as shown below:

P1V1 /T1 = P2V2 /T2

P × V/T = 4P × V2/2T

Cross multiply

T × 4P × V2 = P × V × 2T

Divide both side by T × 4P

V2 = (P × V × 2T) / (T × 4P)

V2 = V/2

V2 = ½V

Therefore, the new volume of the sample is halved .

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

i do not get it   at all