Question

calculate the temperature of 4.0 mole of a gas operating 5 DM cube at 3.32​

Answer 1

If gas is assumed to be ideal then

Use ideal gas equation Pv=nRT

T= PV/(nR)

= 3.32×5/4×0.083

= 50k

Temp of gas is 50 k

Answer 2

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n = 4.0 mol

v = 5 dm^3

p = 3.32 bar

R = 0.083 bar dm3 K^–1 mol^–1

From the ideal gas equation, we get

From the ideal gas equation, we getpV = nRT

From the ideal gas equation, we getpV = nRT⇒ T = pV / nR

From the ideal gas equation, we getpV = nRT⇒ T = pV / nR = (3.32x5) / (4x0.083)

From the ideal gas equation, we getpV = nRT⇒ T = pV / nR = (3.32x5) / (4x0.083) = 50K

From the ideal gas equation, we getpV = nRT⇒ T = pV / nR = (3.32x5) / (4x0.083) = 50KHence, the required temperature is 50 K.

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