pressure exerted by 220g of co2 gas in 8.21dm3 vessel at 27°c ?​

Answers 2

Answer:

P= nRT/V

= 5×0.0821×300/8.21

=15atm

Explanation:

Given; V= 8.21dm^3 = 8.21L

T=27°C = 300K

given mass = 220g

n= 220g/44g = 5mole

PV = nRT

P= nRT/V

= 5×0.0821×300/8.2

=15atm

answer img

Given - Mass, volume, temperature

Find - Pressure exerted

Solution - The pressure exerted by carbon dioxide gas as per the given information is - 15 atm.

The solution will be calculated using the formula -

PV = nRT. In the formula, P is pressure (atm), V is the volume (litre), n is the number of moles, R is gas constant (L-atm/mol-K) and T is the temperature (Kelvin). Units are given in brackets.

So, Number of moles = mass/molar mass

Number of moles = 220/44

Number of moles = 5

Volume = 8.21 dm³

Volume = 8.21 litre

Temperature = 27 + 273

Temperature = 300 K

Gas constant R = 0.082 L atm/mol K

Keep the values in the formula to find the value of pressure exerted.

[tex]Pressure = \frac{5*0.082*300}{8.21} [/tex]

[tex]Pressure = \frac{123}{8.21} [/tex]

[tex]Pressure = 14.98 \: or \: 15 \: atm[/tex]

Thus, the pressure exerted by carbon dioxide gas as per the given information is 15 atm.

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