Question

What external pressure must be supplied to compress
2.76 L of a gas at 298K and 0.878 atm to 2.00 L at
298K?​

Answer 1

Answer:

the reaction is carried out in closed vessel, ∆V = 0

W=−P

x ∆V = 0

(ii) if the reaction is carried out in open beaker (external pressure being 1atm)(initial volume is = 0)

Final volume occupied by 0.2 mole ofH

2

at 25°C and 1 atm pressure can be calculated as follows pV = nRT

V = nrt/p

= 0.2 mol x 0.0821 L atm/K/mol x 298K / 1 atm

=4.89 L

Explanation:

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Answer 2

Explanation:

The reaction is carried out in closed vessel,

delta V = 0.

W = -P

X delta V = 0

if the reaction is carried out in open beaker ( external pressure being 1atm )

( initial volume is = 0 )

Final volume occupied by 0.2 mole of Hz at 25°C and 1 atm pressure can be calculated as follows

PV = nRT

V = nRT / P

=0.2 mol × 0.0821 L atm/K/mol × 298 K/atm

V = 4.89 L