Question

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

Answer 1

Answer:

Hello mate✌✌✌

Given,

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.

Hope it helps✌✌✌

Answer 2

Given,n = 4.0 molv = 5 dm^3p = 3.32 barR = 0.083 bar dm3 K^–1 mol^–1

From the ideal gas equation,

we getFrom the ideal gas equation,

we getpV = nRTFrom the ideal gas equation, we getpV = nRT⇒ T = pV / nRFrom 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) = 50KFrom the ideal gas equation,

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