7. a movable piston contains a sample of 680 ml of neon gas with a temperature of -5 degrees c. when the piston is heated the sample expands to a volume of 1.32 l. what is the new temperature of the neon gas?
8. a helium balloon has a volume of 2600 cm3 when the temperature is 21 degrees c. what is the volume of the balloon when it’s placed in a freezer with a temperature of -15 degrees c?
9. the kelvin temperature of sample of 650 cm3 sample of ammonia gas is doubled. what is the new volume of the gas? assume that the pressure stays constant.
10. a movable piston is allowed to cool from 392 degrees f to 104 degrees f. if the initial volume is 105 ml, what will be the new volume?

7. The new temperature of the neon gas is 520 K.

8. The volume of the balloon when it’s placed in a freezer is 2.3 L.

9. The new volume of the gas is 1.25L.

10. The new volume is 0.07 L.

Explanation: Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

   (At constant pressure and number of moles)

= initial volume of neon gas= 680 ml= 0.68 L

= initial temperature =

= final volume of neon gas = 1.32 L

= final temperature = ?

8.

= initial volume of helium gas = 2600 ml = 2.6L

= initial temperature helium gas =

= final volume helium gas=?

= final temperature of helium gas =

9.

= initial volume of ammonia gas = 650 ml = 0.65L

= initial temperature helium gas =T K

= final volume helium gas=[ ?

= final temperature of helium gas =

10.  

= initial volume =105 ml = 0.105 L

= initial temperature helium gas =

= final volume helium gas=  ?

= final temperature of helium gas =

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