A flexible vessel contains 32.00 L of gas at a pressure of 1.59 atm. Under conditions of constant temperature and moles of gas, what is the volume of the gas when the pressure of the vessel is decreased by a factor of three

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

Explanation:

Step 1: Data given

Volume of the gas = 32.00 L

Pressure of the gas = 1.59 atm

Temperature and moles of gas stay constant

the pressure of the vessel is decreased by a factor of three

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.59 atm

⇒with V1 = the initial volume = 32.00 L

⇒with P2 = the decreased pressure = 1.59 / 3 = 0.53 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.59 atm * 32.00 L = 0.53 atm * V2

V2 = (1.59 * 32.00 L) / 0.53 atm

V2 = 96.00 L

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

1 a

2 d

3 a

4 c

explanation:

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