Question

Find the temperature in °c at which volume and pressure of 1 mole of
nitrogen gas becomes 10 dm³and 2.46 atmosphere respectively.
​

Answer 1

The temperature in °C at which volume and pressure of 1 mole of  nitrogen gas becomes 10 dm³and 2.46 atmosphere respectively is $27^0C$

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 2.46 atm

V= Volume of the gas = $10dm^3=10L$   $1dm^3=1L$

T= Temperature of the gas in Kelvin =?

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 1 mole

$T=\frac{PV}{nR}=\frac{2.46atm\times 10L}{0.0821atmL/Kmol\times 1mol}=300K=(300-273)^0C=27^0C$

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Answer 2

Answer:

27 degree Celsius aprox / 26.63 degree Celsius

Explanation:

pv=nrt

t=pv/nr

= 2.46×10/1×0.821

=300kelvin

conversation

kelvin to degree

300-273

27 degree Celsius

now why I have taken 0.821

because of pressure is in atmosphere