Gas law problem: a gas has a volume of 800 ml at -25℃ and 600 kpa. what would the volume of the gas be at 227℃ and 1000 kpa?

= 930.23 mL

Using the combined gas law;

P1V1/T1 = P2V2/T2

Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and

V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K

Thus;

V2 = P1V1T2/T1P2

     = (600 ×800 ×500) / (258 × 1000)

     = 930.23 mL

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer is c. nacl