A 10.00 g sample of He is compressed isothermally from a volume of 61.1 L to a volume 2.00 L at a temperature of 25 °C.

Calculate the work done by the gas on the surroundings if the external pressure is 101.325 kPa.

25 cm of 1.0 mol dm− NaOH is added to 25 cm of 1.0 mol dm− HCl. The temperature rise is 6¢XC.

Which reactants will also give a temperature rise of 6¢XC?

A. 25 cm of 2.0 mol dm− NaOH and 25 cm of 2.0 mol dm− HCl.
B. 50 cm of 1.0 mol dm− NaOH and 50 cm of 1.0 mol dm− HCl.
C. 50 cm of 0.5 mol dm− NaOH and 50 cm of 0.5 mol dm− HCl.
D. 100 cm of 0.25 mol dm− NaOH and 100 cm of 0.25 mol dm− HCl.

To calculate the work done by the gas on the surroundings during an isothermal compression, you can use the formula:

W = - Pext * ΔV

Where:
W is the work done by the gas on the surroundings,
Pext is the external pressure, and
ΔV is the change in volume.

In this case, the external pressure is given as 101.325 kPa. However, the volume change needs to be in the same units as the pressure, so we need to convert the volume from liters to kiloliters.

Given:
Initial volume (Vi) = 61.1 L
Final volume (Vf) = 2.00 L
External pressure (Pext) = 101.325 kPa

Converting the initial volume to kiloliters:
Vi = 61.1 L * (1 kL / 1000 L)
Vi = 0.0611 kL

Converting the final volume to kiloliters:
Vf = 2.00 L * (1 kL / 1000 L)
Vf = 0.002 kL

Now we can calculate the change in volume:
ΔV = Vf - Vi
ΔV = 0.002 kL - 0.0611 kL
ΔV = -0.0591 kL

Finally, we can substitute the values into the formula to calculate the work done:

W = - Pext * ΔV
W = - (101.325 kPa) * (-0.0591 kL)
W = 5.99 kJ (rounded to two decimal places)

Therefore, the work done by the gas on the surroundings is approximately 5.99 kJ.