Find the temperature at which volume and pressure of $28$ g Nitrogen gas become $10 d m^{3}$ and $2.46$ atm respectively.

1200 K

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600 K

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300 K

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900 K

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Solution

The correct option is C 300 K
Number of moles of nitrogen gas = 28/28 = 1 mol
PV = nRT
T = PV/nR = (2.46 atm x 10L) / 1 x 0.082 L atm / (mol K)
= 300 K

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Find the temperature at which volume and pressure of 28g Nitrogen gas become 10dm3 and 2.46 atm respectively.

The correct option is C 300 KNumber of moles of nitrogen gas = 28/28 = 1 molPV = nRTT = PV/nR = (2.46 atm x 10L) / 1 x 0.082 L atm / (mol K) = 300 K

Find the temperature at which volume and pressure of $28$ g Nitrogen gas become $10 d m^{3}$ and $2.46$ atm respectively.

The correct option is C 300 K
Number of moles of nitrogen gas = 28/28 = 1 mol
PV = nRT
T = PV/nR = (2.46 atm x 10L) / 1 x 0.082 L atm / (mol K)
= 300 K