Apiston chamber filled with ideal gas is kept in a constant-temperature bath at 25.0°c. the piston expands from 25.0 ml to 75.0 ml very, very slowly, as illustrated in the figure below. if there is 0.00100 mole of ideal gas in the chamber, calculate the work done by the system

Answer : The work done by the system is, 2.2722 J

Explanation :

The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 0.00100 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  =

= initial volume of gas  = 25 mL

= final volume of gas  = 75 mL

Now put all the given values in the above formula, we get:

Therefore, the work done by the system is, 2.2722 J

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