Question

pressure exerted by 16g of oxygen gas in 0.01 m^3 vessel at 47 degree celcius is?​

Answer 1

Given - Weight of oxygen gas - 16 gram

Volume of vessel - 0.01 m³

Temperature - 47° C

Find - Pressure exerted.

Solution - Exerted pressure can be calculated by the help of ideal gas law equation - PV = nRT. In the formula, P represents pressure, V is volume, n is number of moles, R is gas constant and T is temperature.

Number of moles = weight/mol weight

Number of moles = 16/32

Number of moles = 0.5

Temperature = 47 + 273 = 320 K

R = 8.2 m³ atm mol-¹ K-¹

Keeping the values in equation-

P = nRT/V

P = 0.5*8.2*320/0.01

P = 131,200 atm

Hence, the exerted pressure will be 131,200 atm.

Answer 2

Given:

16g of oxygen gas is kept in 0.01 m³ vessel at 47°C.

To find:

Pressure exerted by gas ?

Calculation:

• Let's assume oxygen to be an IDEAL GAS in this case. So, we can say that:

$\rm \: PV = nRT$

• 'P' is pressure, 'V' is volume, 'n' is moles, 'R' is Universal Gas Constant and 'T' is temperature.

$\rm \implies\: P \times 0.01 = \dfrac{16}{32} \times 8.2 \times (47 + 273)$

$\rm \implies\: P \times 0.01 = 0.5 \times 8.314\times 320$

$\rm \implies\: P \times 0.01 =1330$

$\rm \implies\: P =133000 \: atm$

So, pressure exerted by oxygen is 133000 atm.