The pressure of a gas in a tank is 3.20 ATM at 22.0°C if the temperature is increased to 60.0°C what will the new pressure in the tank be

P₂ = 3.61 atm

Explanation:

Given data:

Initial pressure of tank = 3.20 atm

Initial temperature = 22.0°C

Final temperature = 60.0°C

Final pressure = ?

Solution:

Initial temperature = 22.0°C (22.0 +273 = 295 K)

Final temperature = 60.0°C (60 +273 = 333)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

3.20 atm / 295 K = P₂/333 K

P₂ = 3.20 atm × 333 K / 295 K

P₂ = 1065.6 atm. K /295 K

P₂ = 3.61 atm

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the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.