Question

At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.

Answer 1

Answer: $106^0C$

Explanation: IDEAL GAS LAW

$PV=nRT$

where,

P = pressure of the gas  = 889 torr = 1.17 atm    (1 torr=0.0013atm)

V = volume of the gas  = 11.8 L

T = temperature of the gas  = ? K

n = number of moles of gas  = 0.444

R = Gas constant = 0.0821 Latm/moleK

$1.17\times 11.8=0.444\times 0.0821\times T$

$T=379K$

$T=(379-273)^0C=106^0C$

Thus temperature of gas is $106^0C$

Answer 2

ideal gas law is: PV = nRT P = pressure (torr) = 889 torr V = volume (Liters) = 11.8 L n = moles of gas = 0.444 mol R = gas constant = 62.4 (L * torr / mol * k) solve for T (in kelvin) T = PV/nR T = (889*11.8)/(.444*62.4) T = 378.6 K convert to C (subtract 273) T = 105.6 deg C