Question

The gas in an engine cylinder is compressed until it has a volume of 0.045 liter and an absolute pressure of 28 atmospheres. if the engine cylinder contains 0.020 mole of gas, what is its temperature
the temperature of the gas is ___ K.

PV=nRT​

Answer 1

Answer:

The temperature of the gas is 767.36 Kelvins

Explanation:

Using ideal gas equation:

PV = nRT

$T=\frac{PV}{nR}$

where,

P = Pressure of gas in an engine= $28 atm$

V = Volume of gas in an engine= 0.045 L

n = number of moles of gas in an engine= 0.020 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas in an engine= ?

Putting values in above equation, we get:

$T=\frac{28 atm\times 0.045 L}{0.020 mol\times 0.0821 L.atm/mol.K}$

T = 767.36 Kelvins

The temperature of the gas is 767.36 Kelvins

Answer 2

Answer:

The temperature of the gas is 767.7 Kelvin

Explanation:

Step 1: Data given

The volume of the gas = 0.045 L

Pressure = 28 atm

Number of moles = 0.020 moles of gas

Step 2: Calculate the temperature of the gas

p*V = n*R*T

T = (p*V)/(n*R)

⇒ with p = the pressure = 28 atm

⇒ with V = the volume of the gas = 0.045 L

⇒ with n = the number of moles of gas = 0.020 moles

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

T = (28*0.045)/(0.020*0.08206)

T = 767.7 K

The temperature of the gas is 767.7 Kelvin