What is the temperature (°c) of 2.48 moles of gas stored in a 30.0 l container at 1559 mm hg? (r= 0.0821 l atm/ mol k)?

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Dejanras Dejanras · Answer 1 · 2018-12-01T16:19:08+01:00

Answer is: the temperature is 28.9°C.

n = 2.48; amount of the gas.

V = 30.0 L; volume of the gas.

p = 1559 mmHg ÷ 760 mmHg/atm.

p = 2.05 atm; pressure of the gas.

R = 0.0821 L·atm/mol·K.

Ideal gas law: p·V = n·R·T.

T = p·V ÷ (n·R).

T = 302.05 K - 273.15.

T = 28.9°C; temperature of the gas.

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mahima6824soni mahima6824soni · Answer 2 · 2022-04-20T11:42:04+02:00

The temperature of 2.48 moles of gas stored in a 30.0 l container at, 1559 mm Hg is 28.9 °C.

Temperature is the measurement of the hotness or coldness of the body. It is denoted by Kelvin.

By the ideal gas law

PV = nrT

[tex]T = \dfrac{PV}{nR}[/tex]

where P = 1559 mm Hg

V = 30.0 l

n = 2.48 moles

r= 0.0821 l atm/ mol k

Putting the values

[tex]T = \dfrac{1559 \times 30.0 l}{2.48 \times 0.0821} = 28.9 ^\circ C[/tex]

Thus, the temperature is 28.9 °C.

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