At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liters, what must the temperature be for the gas pressure to remain constant? A. 160 K B. 273 K C. 560 K D. 140 K

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At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liters, what must the temperature be for the gas pressure to remain constant? A. 160 K B. 273 K C. 560 K D. 140 K

Mathematics
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Is there a formula given
according to this https://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text9/Tx95/tx95.html
Charles' Law states: If the pressure of a gas sample is kept constant, the volume of the sample will vary directly with the temperature in Kelvin (Figure 9.9).

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this means T = k * V 320 = k * 40 k = 320/40 = 8 T = 8*V Can you solve it now?
pv=nRT
P1*V1=n*R*320K P2*V2=n*R*T2 P2=P1 and V2=20.0 or 1/2 of V1 so P2*V2= P1*1/2V1=1/2(P1*V1)=1/2( n*R*320K)= n*R*160K therefore 160K

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