# A sample of an ideal gas has a pressure of 6.50 atm at -15.0 degrees Celsius. At what temperature would the pressure be 75.0 atm?

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A sample of an ideal gas is at a pressure of 6.5 atm at -15 degree Celsius.

The pressure (P), volume (V), temperature (T) and number of moles (n) of an ideal gas are related by P*V = n*R*T where R is a constant.

It is assumed that the volume available for the sample of ideal gas and the amount of the gas does not change as the temperature is changed. Pressure is directly proportional to temperature. The temperature can be converted to Kelvin to change the sign to positive.

-15 C = 273 - 15 = 258 K

6.5 : 258 = 75 : T

=> T = 258*75/6.5 = 2976.92 K or 2703.92 degree Celsius.

The pressure of the sample of gas is 75 atm. at 2703.92 C.

the given data is

P1 = 6.50 atm.

P2 = 75.0 atm.

T1= -15 C=273 - 15 = 258 k

T2 = ? K

according to Gay Lussac's law,

P1/T1=P2/T2 (substuting values)

6.50/258=75.0/T2

T2=75*258*100/650

T2=2976.92

T2=2976.92-273=2703.92 C

at 2703.92 C the pressure becomes equal to 75 atm.