At a given temperature the pressure of a gas reduces to 75%of its initial value and the volume increases by 40% of its initial value.find this temperature of the initial temperature was-10°C
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P1 = P P2 = 75% of P
= 3P/4
V1 = V V2 = V + 40% of V
= 7V/5
T1 = 10°C. T2 = ?
= 273 + 10
= 283 k
By Gas Equation,
(P1 * V1)/T1 = (P2 * V2)/T2
(P * V)/283 = (3P/4 * 7V/5)/T2
PV/283 = 21PV/(20*T2)
→ T2 = 21PV * 283/20PV
T2 = 297.5 k
In °C,
297.5-273
=24.5°C
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= 3P/4
V1 = V V2 = V + 40% of V
= 7V/5
T1 = 10°C. T2 = ?
= 273 + 10
= 283 k
By Gas Equation,
(P1 * V1)/T1 = (P2 * V2)/T2
(P * V)/283 = (3P/4 * 7V/5)/T2
PV/283 = 21PV/(20*T2)
→ T2 = 21PV * 283/20PV
T2 = 297.5 k
In °C,
297.5-273
=24.5°C
I am always ready to help you.
Please follow me.
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